Solved Examples

Before You Start: How to Use These 40 Problems

Sections 1 to 7 taught you the physics. This section is where you find out whether you actually own it. Every problem here is new — new numbers, new framings, and combinations that no single earlier section could set up on its own. Nothing repeats an in-section example you have already worked.

Here is the thing about this chapter: it is not conceptually hard, but it is geometrically unforgiving. Almost every mark lost in Motion in a Plane goes to one of five places — adding magnitudes instead of components, trusting a bare $\tan^{-1}$ and landing in the wrong quadrant, mixing up which subscript is the observer in $\vec{v}_{AB}$, forgetting that $v_y = 0$ at the top of a projectile but $v_x$ is not, or thinking that constant speed means zero acceleration. So work these with a pen, and draw the diagram before you write the first equation. Every single time.

The order they come in

Examples	What they drill	Feels like
1-6	Scalars vs vectors, meaningful operations, displacement vs path length	Warm-up
7-13	Resultants and directions, components, the quadrant trap, vector inequalities, closure	Warm-up to medium
14-19	Differentiating $\vec{r}(t)$, constant acceleration in a plane, the angle between $\vec{v}$ and $\vec{a}$	Medium
20-27	Projectiles: full anatomy, complementary angles, horizontal projection, finding $\theta_0$ from $H$ and $R$, clearing an obstacle	Medium, exam bread-and-butter
28-32	Circular motion: $a_c$ from $v$, from $\omega$, from $T$; period and frequency; comparisons	Medium
33-37	Relative velocity in a plane: rain, both river routes, aircraft in wind, a two-car collision	Medium to hard
38-40	Multi-concept chains — two or three skills locked together	Hard, JEE Advanced flavour

Key Point: Throughout this section, $v_0$ and $\theta_0$ mean the launch speed and launch angle of a projectile. Every problem states the value of $g$ it uses, because this chapter mixes 9.8 m/s^2 with 10 m/s^2 (used wherever a question is built on round numbers). Never carry one over into the other.

[Board Important] In Board physics the setup carries as many marks as the answer: axes stated, components written, formula named, substitution shown. Every solution below is written the way yours should look.

Solved Examples

Example 1: Which of these operations even make sense?

State with reasons whether each of the following algebraic operations with scalar and vector physical quantities is meaningful: (a) adding any two scalars, (b) adding a scalar to a vector of the same dimensions, (c) multiplying any vector by any scalar, (d) multiplying any two scalars, (e) adding any two vectors, (f) adding a component of a vector to the same vector.

Solution:

1. (a) Adding any two scalars — meaningful only if they are of the same dimension. You may add 5 kg to 3 kg. You may not add 5 kg to 3 s. "Any two" is too generous, so the honest answer is: meaningful only for like quantities.
2. (b) Adding a scalar to a vector — never meaningful, even when the dimensions match. A vector carries direction; a scalar does not. There is no rule that tells you which way the sum points. Force 5 N added to work 5 J is nonsense twice over; even "5 N of force plus 5 N of some scalar" has no defined direction.
3. (c) Multiplying a vector by a scalar — always meaningful. This is Section 2's $\lambda\vec{A}$. The scalar may even carry its own units: velocity (a vector) times mass (a scalar) gives momentum (a vector).
4. (d) Multiplying any two scalars — always meaningful. Mass times specific heat, density times volume, current times time. The product simply carries the product of the units.
5. (e) Adding any two vectors — meaningful only if they are of the same dimension. You may add two forces by the triangle law; you may not add a force to a velocity.
6. (f) Adding a component of a vector to the same vector — meaningful. A component of a vector is a vector of the same kind, with the same dimensions. So $\vec{A} + A_x\hat{i}$ is a perfectly legal vector sum. It is unusual, but it is legal.

Final Answer: (a) meaningful only for same-dimension scalars; (b) never meaningful; (c) always meaningful; (d) always meaningful; (e) meaningful only for same-dimension vectors; (f) meaningful.

Takeaway: Two separate filters are running here. Dimensions decide whether an addition is allowed at all; type (scalar vs vector) decides whether the operation itself exists. Multiplication is the relaxed one — you can multiply almost anything by almost anything, because the units simply combine. Addition is the strict one.

Example 2: Five statements, true or false

State with reasons whether each is true or false: (a) the magnitude of a vector is always a scalar, (b) each component of a vector is always a scalar, (c) the total path length is always equal to the magnitude of the displacement vector, (d) the average speed of a particle is greater than or equal to the magnitude of its average velocity over the same interval, (e) three vectors not lying in a plane can never add up to give a null vector.

Solution:

1. (a) TRUE. Magnitude is a single non-negative number with units and no direction. That is the definition of a scalar.
2. (b) FALSE — and this is the one most students get wrong. A component of a vector is itself a vector: $A_x\hat{i}$ is a vector pointing along $\hat{i}$. What is scalar is the coefficient $A_x$. Careful writing distinguishes "the component $A_x\hat{i}$" from "the number $A_x$", and this statement fails on the first reading.
3. (c) FALSE. They are equal only when the particle moves in a straight line without reversing. In every other case the path is longer than the straight-line gap between the ends. A full circle is the extreme case: path length $2\pi R$, displacement zero.
4. (d) TRUE. Average speed is $\dfrac{\text{path length}}{\Delta t}$ and the magnitude of average velocity is $\dfrac{|\Delta\vec{r}|}{\Delta t}$. Since path length $\ge |\Delta\vec{r}|$ always, and both carry the same $\Delta t$, the inequality follows immediately.
5. (e) TRUE. Adding three vectors head to tail gives a closed triangle only if the three of them are coplanar — a triangle is a plane figure. If the third vector sticks out of the plane of the first two, the tail-to-head chain cannot close. You need at least four non-coplanar vectors to sum to zero in three dimensions.

Final Answer: (a) T, (b) F, (c) F, (d) T, (e) T.

Takeaway: Statement (b) is the trap and statement (d) is the one worth remembering as a principle: average speed $\ge$ magnitude of average velocity, always, with equality only for straight-line one-way motion. It follows from nothing more than "path length is never shorter than the shortcut".

Example 3: What actually makes a quantity a scalar?

Read each statement and state, with reasons and examples, whether it is true or false. A scalar quantity is one that (a) is conserved in a process, (b) can never take negative values, (c) must be dimensionless, (d) does not vary from one point to another in space, (e) has the same value for observers with different orientations of axes.

Solution:

1. (a) FALSE. Conservation has nothing to do with being a scalar. Energy is a conserved scalar, but kinetic energy alone is a scalar that is not conserved in an inelastic collision. Meanwhile momentum is conserved and is a vector. The two ideas are independent.
2. (b) FALSE. Temperature on the Celsius scale is a scalar and reads $-10$°C on a cold morning. Potential energy, electric potential and charge are all scalars that go negative freely.
3. (c) FALSE. Mass is a scalar with dimension $[M]$; speed is a scalar with dimension $[LT^{-1}]$. Most scalars in physics carry units. (Refractive index and strain are the dimensionless ones — but that is a coincidence, not a requirement.)
4. (d) FALSE. Temperature varies from point to point in a room, gravitational potential varies with height, and density varies through the atmosphere. All three are scalars. Varying in space is perfectly normal.
5. (e) TRUE — and this is the real definition. A scalar is exactly the thing that is invariant under rotation of the coordinate axes. Rotate your axes and $A_x$ and $A_y$ both change, but $|\vec{A}| = \sqrt{A_x^2 + A_y^2}$ does not. Mass, temperature and speed are the same for every observer no matter how his axes are tilted.

Final Answer: (a) F, (b) F, (c) F, (d) F, (e) T.

Takeaway: Four tempting definitions of "scalar" are all wrong and only the fifth survives. A scalar is a quantity that does not change when you rotate the axes. Everything else — conservation, sign, dimensions, uniformity in space — is a separate property that some scalars happen to have and others do not.

Example 4: Three skaters, one displacement

Three girls skating on a circular ice ground of radius 200 m start from a point P on the edge and reach the diametrically opposite point Q, each following a different path — one along the diameter, one along a curved path bowing to one side, one along a semicircular arc. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of path skated?

Solution:

1. Set coordinates. Put the centre at the origin, so P is at $(-200, 0)$ m and Q at $(+200, 0)$ m.
2. Displacement is end minus start, and nothing else about the path enters: $$\Delta\vec{r} = \vec{r}_Q - \vec{r}_P = (200 - (-200))\hat{i} = 400\hat{i}\ \text{m}$$
3. So the magnitude is 400 m for all three girls, directed from P to Q. The three paths are different; the three displacements are identical.
4. Now the path lengths. The straight diameter is 400 m. The semicircular arc is $\pi R = \pi(200) = 628.3$ m. The bowed curve is somewhere in between, but certainly more than 400 m.
5. The match. Path length equals displacement magnitude only for the girl who skated along the straight diameter, because she never changed direction.

Final Answer: All three have displacement of magnitude 400 m directed from P to Q. Only the girl who skates along the diameter has path length equal to it (400 m).

Takeaway: Displacement is blind to the route. Three completely different journeys with the same endpoints give one identical displacement vector, and the straight-line path is the only one whose length matches it — which is another way of saying a straight line is the shortest distance between two points.

Example 5: The cyclist in the park

A cyclist starts from the centre O of a circular park of radius 1 km, rides out to the edge at P, then cycles along the circumference through a quarter of the circle to Q, and returns to the centre along QO. The round trip takes 10 minutes. Find (a) the net displacement, (b) the average velocity, and (c) the average speed.

Solution:

1. Recognise the shape of the journey. He starts at O and finishes at O. Everything else is decoration.
2. (a) Net displacement. Initial and final position vectors are the same point, so $$\Delta\vec{r} = \vec{r}_{\text{final}} - \vec{r}_{\text{initial}} = \vec{0}$$ The net displacement is the null vector — zero magnitude, no direction to speak of.
3. (b) Average velocity $= \dfrac{\Delta\vec{r}}{\Delta t} = \dfrac{\vec{0}}{10\ \text{min}} = \vec{0}$. Zero, without computing anything.
4. (c) Average speed needs the path length, and here the route matters completely:

• O to P along a radius: 1 km
• P to Q along a quarter of the circumference: $\dfrac{2\pi R}{4} = \dfrac{\pi(1)}{2} = 1.571$ km
• Q back to O along a radius: 1 km $$\text{path length} = 1 + 1.571 + 1 = 3.571\ \text{km}$$

1. Convert the time and divide. $10$ min $= \dfrac{1}{6}$ h, so $$\text{average speed} = \frac{3.571}{1/6} = 21.4\ \text{km/h}$$

Final Answer: (a) zero, (b) zero, (c) 21.4 km/h.

Takeaway: For any closed journey the average velocity is zero before you compute anything — so if a question asks for it, write zero and move on. The average speed is the quantity that carries information, and it is the only one of the two that needs you to look at the actual route.

Example 6: The dishonest cabman

A passenger wants to go from the station to a hotel 10 km away on a straight road. A dishonest cabman takes him along a circuitous 23 km route and reaches the hotel in 28 minutes. Find (a) the average speed of the taxi and (b) the magnitude of the average velocity. Are the two equal?

Solution:

1. Sort the two lengths. Path length is what the meter records: 23 km. Displacement is what the map shows: 10 km, straight from station to hotel.
2. Time in hours. $28$ min $= \dfrac{28}{60} = 0.4667$ h.
3. (a) Average speed $= \dfrac{\text{path length}}{\Delta t} = \dfrac{23}{0.4667} = 49.3$ km/h.
4. (b) Magnitude of average velocity $= \dfrac{|\Delta\vec{r}|}{\Delta t} = \dfrac{10}{0.4667} = 21.4$ km/h.
5. (c) Not equal, and they could not be: the route was not straight. Their ratio is $$\frac{49.3}{21.4} = \frac{23}{10} = 2.3$$ which is exactly the ratio of path length to displacement — the times cancel.

Final Answer: (a) 49.3 km/h, (b) 21.4 km/h, (c) no, they are not equal; the speed is 2.3 times the velocity magnitude.

Takeaway: Notice you never needed the shape of the circuitous route. The ratio of average speed to the magnitude of average velocity is always just the ratio of path length to displacement magnitude, because the same $\Delta t$ divides both. That single sentence answers a whole family of questions in one line.

Example 7: Two forces at an angle — magnitude and direction

Two forces of magnitude 12 N and 16 N act at a point with an angle of $60°$ between them. Find the magnitude of the resultant and the angle it makes with the 12 N force. Solve it twice: once with the resultant formula and once by components.

Solution:

1. Route 1 — the resultant formula from Section 3. With $A = 12$ N, $B = 16$ N, $\theta = 60°$: $$R = \sqrt{A^2 + B^2 + 2AB\cos\theta} = \sqrt{144 + 256 + 2(12)(16)(0.5)}$$ $$R = \sqrt{144 + 256 + 192} = \sqrt{592} = 24.3\ \text{N}$$
2. Its direction from the 12 N force: $$\tan\alpha = \frac{B\sin\theta}{A + B\cos\theta} = \frac{16(0.8660)}{12 + 16(0.5)} = \frac{13.86}{20} = 0.6928$$ $$\alpha = \tan^{-1}(0.6928) = 34.7°$$
3. Route 2 — components, as a check. Put $\vec{A}$ along $+x$: $$\vec{A} = 12\hat{i}, \qquad \vec{B} = 16\cos 60°\,\hat{i} + 16\sin 60°\,\hat{j} = 8\hat{i} + 13.86\hat{j}$$
4. Add componentwise: $$\vec{R} = (12 + 8)\hat{i} + (0 + 13.86)\hat{j} = 20\hat{i} + 13.86\hat{j}$$ $$R = \sqrt{20^2 + 13.86^2} = \sqrt{400 + 192} = \sqrt{592} = 24.3\ \text{N}\ \checkmark$$
5. Direction from components: $\tan\alpha = \dfrac{13.86}{20}$, giving $\alpha = 34.7°$ above the 12 N force. Both components are positive, so the resultant is in the first quadrant and the plain $\tan^{-1}$ is safe here. $\checkmark$

Final Answer: $R = 24.3$ N at $34.7°$ from the 12 N force, on the side of the 16 N force.

Takeaway: The two routes are the same algebra wearing different clothes — expand $|\vec{A}+\vec{B}|^2$ in components and the $2AB\cos\theta$ falls out. Use the formula when only two vectors are involved; switch to components the moment there are three or more, because the formula does not extend and components do.

Example 8: Four vectors, and the quadrant that catches everyone

Find the resultant of the four displacement vectors $$\vec{A} = 8\hat{i} + 6\hat{j}, \quad \vec{B} = -5\hat{i} + 2\hat{j}, \quad \vec{C} = -9\hat{i} - 7\hat{j}, \quad \vec{D} = -2\hat{i} - 9\hat{j}$$ all in metres. Give its magnitude and its true direction. What does a calculator's $\tan^{-1}$ give, and why is that wrong? What single vector would have to be added to make the resultant zero?

Solution:

1. Add the $x$-components and the $y$-components separately. This is the only way to add more than two vectors — the $R = \sqrt{A^2+B^2+2AB\cos\theta}$ formula handles exactly two and does not extend. $$R_x = 8 - 5 - 9 - 2 = -8\ \text{m}$$ $$R_y = 6 + 2 - 7 - 9 = -8\ \text{m}$$ $$\vec{R} = -8\hat{i} - 8\hat{j}\ \text{m}$$
2. Magnitude: $$|\vec{R}| = \sqrt{(-8)^2 + (-8)^2} = \sqrt{128} = 8\sqrt{2} = 11.3\ \text{m}$$
3. Direction — and here is the trap. Both components are negative, so the resultant lies in the third quadrant. The reference angle from the magnitudes alone is $$\tan^{-1}\!\left(\frac{8}{8}\right) = 45°$$ so the true direction is $$180° + 45° = 225°\ \ \text{anticlockwise from}\ +x$$ (equivalently $-135°$, or "$45°$ below the negative $x$-axis").
4. What the calculator says. Fed $\tan^{-1}(R_y/R_x) = \tan^{-1}(-8/-8) = \tan^{-1}(1)$, it returns $45°$ — pointing into the first quadrant, exactly $180°$ from the truth. The two minus signs cancelled inside the ratio and the calculator lost the information forever.
5. A sanity check on the answer. The four vectors have magnitudes 10, 5.39, 11.40 and 9.22 m, summing to 36.0 m. The resultant is 11.3 m — comfortably below that ceiling, as the triangle inequality demands. And a resultant pointing down and to the left is what you would expect from vectors $\vec{C}$ and $\vec{D}$, which are the two big ones and both point that way.
6. The vector that cancels it. To make the total zero we need $\vec{E}$ with $\vec{R} + \vec{E} = \vec{0}$: $$\vec{E} = -\vec{R} = 8\hat{i} + 8\hat{j}\ \text{m}$$ which has the same magnitude 11.3 m but points at $45°$ — the exact opposite direction.

Final Answer: $\vec{R} = (-8\hat{i} - 8\hat{j})$ m, of magnitude 11.3 m at $225°$ from the $+x$ axis. The calculator's $45°$ is wrong by $180°$. Adding $\vec{E} = (8\hat{i} + 8\hat{j})$ m makes the resultant zero.

Takeaway: Two habits, both non-negotiable. Add components, never magnitudes — and for three or more vectors that is the only method available. Then fix the quadrant from the signs, because $\tan^{-1}$ can only ever return an answer between $-90°$ and $+90°$ and physically half of all directions live outside that window. [JEE/NEET] A resultant with two negative components is always in the third quadrant, whatever your calculator insists.

Example 9: The four inequalities, established and tested

Establish the inequalities (a) $|\vec{a}+\vec{b}| \le |\vec{a}| + |\vec{b}|$, (b) $|\vec{a}+\vec{b}| \ge \big||\vec{a}| - |\vec{b}|\big|$, (c) $|\vec{a}-\vec{b}| \le |\vec{a}| + |\vec{b}|$, (d) $|\vec{a}-\vec{b}| \ge \big||\vec{a}| - |\vec{b}|\big|$. When does the equality sign apply? Then test all four with $|\vec{a}| = 5$, $|\vec{b}| = 3$ at $60°$.

Solution:

1. The geometry behind (a) and (b). Draw $\vec{a}$ and $\vec{b}$ head to tail; $\vec{a}+\vec{b}$ closes the triangle. In any triangle no side can exceed the sum of the other two and no side can be less than their difference. Those two facts are inequalities (a) and (b).
2. Equality in (a) needs the triangle to collapse into a straight line with $\vec{a}$ and $\vec{b}$ pointing the same way ($\theta = 0$). Equality in (b) needs them antiparallel ($\theta = 180°$).
3. (c) and (d) come free. Write $\vec{a} - \vec{b} = \vec{a} + (-\vec{b})$. Since $|-\vec{b}| = |\vec{b}|$, apply (a) and (b) to the pair $\vec{a}$ and $-\vec{b}$. The bounds are unchanged; only the equality conditions swap over, because reversing $\vec{b}$ swaps "parallel" and "antiparallel". So (c) is an equality when $\vec{a}$ and $\vec{b}$ are antiparallel and (d) when they are parallel.
4. The algebraic check. For any angle $\theta$, $$|\vec{a} \pm \vec{b}|^2 = a^2 + b^2 \pm 2ab\cos\theta$$ Since $\cos\theta$ can only run from $-1$ to $+1$, both expressions are trapped between $(a-b)^2$ and $(a+b)^2$. That is all four inequalities in one line.
5. The numbers. With $a = 5$, $b = 3$, the bounds are $|5 - 3| = 2$ and $5 + 3 = 8$. At $\theta = 60°$: $$|\vec{a}+\vec{b}| = \sqrt{25 + 9 + 2(15)(0.5)} = \sqrt{49} = 7.00$$ $$|\vec{a}-\vec{b}| = \sqrt{25 + 9 - 2(15)(0.5)} = \sqrt{19} = 4.36$$ Both sit comfortably inside $[2, 8]$. $\checkmark$

Final Answer: All four follow from the triangle inequality; equality in (a) and (d) when the vectors are parallel, in (b) and (c) when they are antiparallel. At $60°$ the sum is 7.00 and the difference 4.36, both inside the bounds 2 to 8.

Takeaway: Memorise the band, not the four separate statements: both $|\vec{a}+\vec{b}|$ and $|\vec{a}-\vec{b}|$ live between $|a-b|$ and $a+b$. Any exam option outside that band is wrong on sight, and this kills a surprising number of multiple-choice questions in ten seconds.

Example 10: Four vectors that sum to zero

Given $\vec{a} + \vec{b} + \vec{c} + \vec{d} = 0$, which of the following are correct? (a) $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ must each be a null vector. (b) The magnitude of $(\vec{a}+\vec{c})$ equals the magnitude of $(\vec{b}+\vec{d})$. (c) The magnitude of $\vec{a}$ can never be greater than the sum of the magnitudes of $\vec{b}, \vec{c}$ and $\vec{d}$. (d) $\vec{b}+\vec{c}$ must lie in the plane of $\vec{a}$ and $\vec{d}$ if $\vec{a}$ and $\vec{d}$ are not collinear, and in the line of $\vec{a}$ and $\vec{d}$ if they are collinear.

Solution:

1. (a) INCORRECT. Four non-zero vectors placed head to tail around a closed quadrilateral sum to zero. A concrete counterexample settles it: take $$\vec{a} = 3\hat{i} + \hat{j}, \quad \vec{b} = -2\hat{i} + 4\hat{j}, \quad \vec{c} = 5\hat{i} - 2\hat{j}, \quad \vec{d} = -6\hat{i} - 3\hat{j}$$ The four components add to $(3-2+5-6, \ 1+4-2-3) = (0, 0)$, yet not one of them is null.
2. (b) CORRECT. Rearranging the given equation, $$(\vec{a}+\vec{c}) = -(\vec{b}+\vec{d})$$ Two vectors that are exact negatives have equal magnitudes (and opposite directions). Check on the numbers above: $\vec{a}+\vec{c} = 8\hat{i} - \hat{j}$ and $\vec{b}+\vec{d} = -8\hat{i} + \hat{j}$, both of magnitude $\sqrt{65} = 8.06$. $\checkmark$
3. (c) CORRECT. From the same equation, $\vec{a} = -(\vec{b}+\vec{c}+\vec{d})$, so $$|\vec{a}| = |\vec{b}+\vec{c}+\vec{d}| \le |\vec{b}| + |\vec{c}| + |\vec{d}|$$ by the triangle inequality applied twice. So $|\vec{a}|$ can equal that sum (if the three are all parallel) but never exceed it.
4. (d) CORRECT. Write $\vec{b}+\vec{c} = -(\vec{a}+\vec{d})$. The vector $\vec{a}+\vec{d}$ is a combination of $\vec{a}$ and $\vec{d}$, so it must lie in the plane those two define. If $\vec{a}$ and $\vec{d}$ happen to be collinear they define no plane, only a line — and then $\vec{a}+\vec{d}$, and hence $\vec{b}+\vec{c}$, lies along that line.

Final Answer: (a) incorrect; (b), (c) and (d) are all correct.

Takeaway: Every part here was solved by the same move — rearrange the zero-sum equation so the thing you are asked about sits alone on one side. Once $(\vec{a}+\vec{c}) = -(\vec{b}+\vec{d})$ is on the page, part (b) is obvious. Do not try to visualise four vectors at once; do the algebra first.

Example 11: The motorist who turns left every 500 m

On open ground a motorist follows a track that turns left by $60°$ after every 500 m. Starting from a given turn, find the displacement at the third, sixth and eighth turn, and compare each with the total path length.

Solution:

1. Recognise the figure. A turn of $60°$ repeated after equal legs is the exterior angle of a regular hexagon: $360°/60° = 6$ legs close the loop exactly. Draw it before computing anything.
2. Set up components. Take the first leg along $+x$. The $k$-th leg is $500$ m in the direction $(k-1)\times 60°$, so its components are $500\cos[(k-1)60°]$ and $500\sin[(k-1)60°]$.
3. At the third turn — add the first three legs: $$\Delta x = 500(\cos 0° + \cos 60° + \cos 120°) = 500(1 + 0.5 - 0.5) = 500\ \text{m}$$ $$\Delta y = 500(\sin 0° + \sin 60° + \sin 120°) = 500(0 + 0.866 + 0.866) = 866\ \text{m}$$ $$|\Delta\vec{r}| = \sqrt{500^2 + 866^2} = \sqrt{1{,}000{,}000} = 1000\ \text{m}$$ Direction: $\tan^{-1}(866/500) = 60°$ to the initial direction (both components positive, first quadrant, so this is safe). Path length $= 3 \times 500 = 1500$ m.
4. At the sixth turn — the hexagon has closed. He is back where he started: $$|\Delta\vec{r}| = 0, \qquad \text{path length} = 6 \times 500 = 3000\ \text{m}$$
5. At the eighth turn — legs 7 and 8 simply retrace legs 1 and 2, so his position is the same as after two legs: $$\Delta x = 500(1 + 0.5) = 750\ \text{m}, \qquad \Delta y = 500(0 + 0.866) = 433\ \text{m}$$ $$|\Delta\vec{r}| = \sqrt{750^2 + 433^2} = 866\ \text{m}\ \ \text{at}\ 30°\ \text{to the initial direction}$$ Path length $= 8 \times 500 = 4000$ m.
6. The comparison.

Turn	Displacement magnitude	Path length	Ratio
3rd	1000 m at $60°$	1500 m	0.67
6th	0	3000 m	0
8th	866 m at $30°$	4000 m	0.22

Final Answer: 1000 m at $60°$ (path 1500 m); zero (path 3000 m); 866 m at $30°$ (path 4000 m).

Takeaway: Look down the last column. The path length climbs steadily while the displacement falls to zero and then climbs back. That is the whole lesson of displacement versus distance in one table: the longer a closed-ish journey runs, the more brutally the two diverge. Also note the shortcut — once you spot the hexagon, the 6th, 12th and 18th turns are all zero without any arithmetic at all.

Example 12: Components along a rotated pair of axes

$\hat{i}$ and $\hat{j}$ are unit vectors along the $x$- and $y$-axes. What are the magnitude and direction of the vectors $\hat{i}+\hat{j}$ and $\hat{i}-\hat{j}$? What are the components of $\vec{A} = 2\hat{i} + 3\hat{j}$ along the directions of $\hat{i}+\hat{j}$ and $\hat{i}-\hat{j}$?

Solution:

1. The vector $\hat{i}+\hat{j}$. Components $(1, 1)$, so $$|\hat{i}+\hat{j}| = \sqrt{1^2 + 1^2} = \sqrt{2} = 1.414$$ Both components positive $\Rightarrow$ first quadrant $\Rightarrow$ direction $\tan^{-1}(1/1) = 45°$ from the $x$-axis.
2. The vector $\hat{i}-\hat{j}$. Components $(1, -1)$, so the magnitude is again $\sqrt{2} = 1.414$. Now $x > 0$, $y < 0$ $\Rightarrow$ fourth quadrant $\Rightarrow$ direction $-45°$ (that is, $45°$ below the $x$-axis, or $315°$).
3. These two are perpendicular to each other ($45°$ and $-45°$ are $90°$ apart), so they are a perfectly good pair of axes — just the usual axes rotated by $45°$. Their unit vectors are $$\hat{n}_1 = \frac{\hat{i}+\hat{j}}{\sqrt{2}}, \qquad \hat{n}_2 = \frac{\hat{i}-\hat{j}}{\sqrt{2}}$$
4. Resolve $\vec{A}$ onto the new axes. $\vec{A} = 2\hat{i}+3\hat{j}$ has magnitude $\sqrt{4+9} = \sqrt{13} = 3.606$ and direction $\tan^{-1}(3/2) = 56.31°$. The component of a vector along any direction is $A\cos(\text{angle between them})$: $$A_1 = 3.606\cos(56.31° - 45°) = 3.606\cos(11.31°) = 3.54$$ $$A_2 = 3.606\cos(56.31° - (-45°)) = 3.606\cos(101.31°) = -0.71$$
5. Cross-check by projecting components directly. Along $\hat{n}_1$ the projection is $\dfrac{2+3}{\sqrt{2}} = \dfrac{5}{1.414} = 3.54$; along $\hat{n}_2$ it is $\dfrac{2-3}{\sqrt{2}} = \dfrac{-1}{1.414} = -0.71$. $\checkmark$
6. Final check on the pair. Rotating axes must not change the length: $$A_1^2 + A_2^2 = (3.54)^2 + (-0.71)^2 = 12.5 + 0.5 = 13 = |\vec{A}|^2\ \checkmark$$

Final Answer: Both $\hat{i}+\hat{j}$ and $\hat{i}-\hat{j}$ have magnitude $\sqrt{2} = 1.414$, directed at $45°$ and $-45°$. The components of $\vec{A}$ are $\dfrac{5}{\sqrt{2}} = 3.54$ along $\hat{i}+\hat{j}$ and $-\dfrac{1}{\sqrt{2}} = -0.71$ along $\hat{i}-\hat{j}$.

Takeaway: Axes are a choice, not a fact. Resolve the same vector onto a rotated pair and both numbers change — but $A_1^2 + A_2^2$ does not, because the length of a vector is a scalar and scalars are exactly the things that survive a rotation of axes (which is precisely what Example 3(e) said). The negative sign on $A_2$ is real information: $\vec{A}$ leans to the $\hat{j}$ side of the $45°$ line.

Example 13: Can these close? Resultant bounds in action

(a) Can three forces of 5 N, 9 N and 17 N acting at a point give zero resultant? (b) Can 3 N, 6 N and 8 N? (c) Two vectors of magnitude 8 and 5 units are added. Which of 2, 3, 9, 13 and 14 units are possible resultants? (d) Show that three forces of 7 N each can give zero resultant, and say how.

Solution:

1. The closure rule. Vectors sum to zero exactly when, drawn head to tail, they close a polygon. For three vectors that polygon is a triangle, so zero resultant is possible if and only if no one magnitude exceeds the sum of the other two.
2. (a) 5, 9 and 17. The two smaller ones give $5 + 9 = 14$, which is less than 17. The 17 N side is too long for the other two to reach across. Not possible. Equivalently: the resultant of 5 and 9 lies between 4 and 14, and 17 is outside that band, so nothing can cancel it.
3. (b) 3, 6 and 8. Check the longest: $3 + 6 = 9 > 8$. $\checkmark$ The triangle closes. Possible.
4. (c) 8 and 5 together. The band of achievable resultants is $$|8 - 5| \le R \le 8 + 5 \quad\Rightarrow\quad 3 \le R \le 13$$ So 3 is possible (antiparallel), 9 is possible (at some intermediate angle), 13 is possible (parallel). 2 is impossible (below the floor) and 14 is impossible (above the ceiling).
5. (d) Three forces of 7 N. Three equal magnitudes always close — into an equilateral triangle. Drawn head to tail that means each successive force is turned $120°$ from the previous one. Check by components with the three at $0°$, $120°$, $240°$: $$\Sigma F_x = 7(1 - 0.5 - 0.5) = 0, \qquad \Sigma F_y = 7(0 + 0.866 - 0.866) = 0\ \checkmark$$

Final Answer: (a) no; (b) yes; (c) 3, 9 and 13 are possible, 2 and 14 are not; (d) yes — with the three forces at $120°$ to one another.

Takeaway: Two rules cover this entire family. Two vectors: the resultant lives in the band $|A-B|$ to $A+B$. Three vectors: they can cancel exactly when the largest is not bigger than the sum of the other two. Neither needs a single trigonometric function, and between them they answer most "is this possible" questions in the exam in under ten seconds.

Example 14: Differentiating a position vector

The position of a particle is given by $\vec{r} = 3.0t\,\hat{i} - 2.0t^2\,\hat{j} + 4.0\,\hat{k}$ metres, where $t$ is in seconds. (a) Find $\vec{v}$ and $\vec{a}$. (b) What are the magnitude and direction of the velocity at $t = 2.0$ s?

Solution:

1. Differentiate component by component. This is the whole method — the $\hat{i}$, $\hat{j}$, $\hat{k}$ are constant vectors, so they just sit there while the coefficients get differentiated: $$\vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}(3.0t)\hat{i} + \frac{d}{dt}(-2.0t^2)\hat{j} + \frac{d}{dt}(4.0)\hat{k}$$ $$\boxed{\vec{v} = 3.0\,\hat{i} - 4.0t\,\hat{j}}\ \ \text{m/s}$$
2. Differentiate once more: $$\vec{a} = \frac{d\vec{v}}{dt} = 0\,\hat{i} - 4.0\,\hat{j} = -4.0\,\hat{j}\ \ \text{m/s}^2$$ The acceleration is constant: 4.0 m/s^2 directed along $-\hat{j}$, for all time.
3. Read what each term is telling you. The $x$-motion is uniform (3.0 m/s forever), the $z$-coordinate never changes (the particle stays in the plane $z = 4.0$ m), and only $y$ accelerates. This is a projectile in disguise.
4. (b) Velocity at $t = 2.0$ s. Substitute: $$\vec{v}(2.0) = 3.0\,\hat{i} - 4.0(2.0)\,\hat{j} = 3.0\,\hat{i} - 8.0\,\hat{j}\ \ \text{m/s}$$
5. Magnitude: $$|\vec{v}| = \sqrt{(3.0)^2 + (-8.0)^2} = \sqrt{9 + 64} = \sqrt{73} = 8.54\ \text{m/s}$$
6. Direction. $v_x > 0$ and $v_y < 0$ puts it in the fourth quadrant of the x-y plane: $$\alpha = \tan^{-1}\!\left(\frac{8.0}{3.0}\right) = 69.4°\ \text{below the}\ +x\ \text{axis}$$ that is, $-69.4°$ measured anticlockwise from $+x$.

Final Answer: (a) $\vec{v} = 3.0\hat{i} - 4.0t\,\hat{j}$ m/s and $\vec{a} = -4.0\hat{j}$ m/s^2. (b) At $t = 2.0$ s the speed is 8.54 m/s directed $69.4°$ below the $x$-axis.

Takeaway: Differentiating a vector is not a new operation — it is ordinary differentiation done once per component. And notice the constant $4.0\hat{k}$ vanished in the very first derivative, taking the entire $z$-motion with it: a constant term in $\vec{r}$ shifts where the motion happens, never how it happens.

Example 15: Constant acceleration from the origin

A particle starts from the origin at $t = 0$ with velocity $10.0\,\hat{j}$ m/s and moves in the x-y plane with constant acceleration $(8.0\hat{i} + 2.0\hat{j})$ m/s^2. (a) At what time is the $x$-coordinate 16 m, and what is the $y$-coordinate then? (b) What is the speed at that time?

Solution:

1. Split into two independent one-dimensional problems. This is the whole point of Section 4. Write the two columns side by side:

	$x$-motion	$y$-motion
initial position	0	0
initial velocity	0	10.0 m/s
acceleration	8.0 m/s^2	2.0 m/s^2

2. (a) Use the $x$-column alone to find the time. With $x_0 = 0$ and $v_{0x} = 0$: $$x = v_{0x}t + \tfrac{1}{2}a_x t^2 = \tfrac{1}{2}(8.0)t^2 = 4.0t^2$$ $$16 = 4.0t^2 \quad\Rightarrow\quad t^2 = 4.0 \quad\Rightarrow\quad t = 2.0\ \text{s}$$ (The root $t = -2.0$ s is before the motion started, so it is rejected.)
3. Now hand that time to the $y$-column — the clock is the only thing the two motions share: $$y = v_{0y}t + \tfrac{1}{2}a_y t^2 = (10.0)(2.0) + \tfrac{1}{2}(2.0)(2.0)^2 = 20 + 4 = 24\ \text{m}$$
4. (b) Velocity at $t = 2.0$ s, from $\vec{v} = \vec{v}_0 + \vec{a}t$: $$v_x = 0 + (8.0)(2.0) = 16\ \text{m/s}, \qquad v_y = 10.0 + (2.0)(2.0) = 14\ \text{m/s}$$
5. Speed: $$|\vec{v}| = \sqrt{16^2 + 14^2} = \sqrt{256 + 196} = \sqrt{452} = 21.3\ \text{m/s}$$

Final Answer: (a) $t = 2.0$ s, at which $y = 24$ m. (b) The speed is 21.3 m/s.

Takeaway: The whole solution is two separate 1D problems that talk to each other through one shared quantity — the time. Solve for $t$ in whichever direction the question gives you enough information, then carry that $t$ across. This is exactly the machinery that makes projectile motion work, and it is worth recognising it here in its pure form.

Example 16: Which relations survive for arbitrary motion?

For any arbitrary motion in space, which of the following are true? (a) $\vec{v}_{avg} = \tfrac{1}{2}[\vec{v}(t_1) + \vec{v}(t_2)]$, (b) $\vec{v}_{avg} = \dfrac{\vec{r}(t_2) - \vec{r}(t_1)}{t_2 - t_1}$, (c) $\vec{v}(t) = \vec{v}(0) + \vec{a}t$, (d) $\vec{r}(t) = \vec{r}(0) + \vec{v}(0)t + \tfrac{1}{2}\vec{a}t^2$, (e) $\vec{a}_{avg} = \dfrac{\vec{v}(t_2) - \vec{v}(t_1)}{t_2 - t_1}$.

Solution:

1. Sort the list into two kinds of statement. Some of these are definitions (true always, by construction); the rest are results derived under an assumption (true only when the assumption holds).
2. (b) TRUE — it is the definition of average velocity. Displacement divided by the time interval. Nothing was assumed about how the particle got from one place to the other.
3. (e) TRUE — it is the definition of average acceleration. Change in velocity divided by the time interval. Again, no assumption.
4. (c) FALSE in general. This came from integrating $\vec{a} = d\vec{v}/dt$ and pulling $\vec{a}$ out of the integral because it was constant. If $\vec{a}$ changes with time you cannot pull it out and the result collapses.
5. (d) FALSE in general, for exactly the same reason — it is the second integration of the same constant-$\vec{a}$ assumption. Also note the symbol $\vec{a}$ in (c) and (d) has no time label, which is itself the giveaway.
6. (a) FALSE in general. Averaging the two end velocities ignores everything in between. It happens to be correct when $\vec{a}$ is constant, because then $\vec{v}$ changes linearly in time and the average of a linear function over an interval is the mean of its endpoint values. Any other motion and it fails. A particle that shoots out and comes straight back to its start has $\vec{v}(t_1) = -\vec{v}(t_2)$ in some cases, giving zero on the right when the true average velocity is genuinely zero — but change the story slightly and the agreement disappears.

Final Answer: Only (b) and (e) are true for arbitrary motion. (a), (c) and (d) require constant acceleration.

Takeaway: Definitions survive everything; derived formulas carry their assumptions with them. Whenever a formula in kinematics contains a bare $\vec{a}$ with no time-dependence written on it, ask immediately: was this derived assuming $\vec{a}$ is constant? For (a), (c) and (d) the answer is yes, which is exactly why they fail here.

Example 17: How fast is that aircraft?

An aircraft flies at a height of 3400 m above the ground. The angle subtended at a ground observation point by the aircraft's positions 10.0 s apart is $30°$. What is the speed of the aircraft?

Solution:

1. Draw the geometry. The observer is at O on the ground. The aircraft is at A at the first instant and at B ten seconds later, both at height 3400 m. The angle $\angle AOB = 30°$.
2. Use the symmetry. Drop a perpendicular from O to the flight line; call its foot M, at the midpoint of AB. This perpendicular is the height, $OM = 3400$ m, and it bisects the $30°$ angle, so $\angle AOM = \angle MOB = 15°$.
3. Solve one right triangle. In triangle OMA, right-angled at M: $$\tan 15° = \frac{AM}{OM} \quad\Rightarrow\quad AM = 3400 \times \tan 15° = 3400 \times 0.2679 = 911.0\ \text{m}$$
4. Double it to get the whole path flown: $$AB = 2 \times AM = 2 \times 911.0 = 1822\ \text{m}$$
5. Divide by the time. The aircraft flew 1822 m in 10.0 s: $$v = \frac{1822}{10.0} = 182\ \text{m/s}$$
6. Sanity check in familiar units. $182 \times 3.6 = 656$ km/h. That is a perfectly ordinary cruising speed for a passenger jet, so the answer is believable.

Final Answer: About 182 m/s (roughly 656 km/h).

Takeaway: The single move that makes this easy is bisecting the angle with the perpendicular from the observer, which turns one awkward isosceles triangle into two clean right-angled ones. Also note the classic trap: it is $\tan 15°$, not $\tan 30°$ — if you use the full angle you get 1961 m/s and an aircraft flying at Mach 6.

Example 18: A full workup with constant acceleration in a plane

A particle is at $\vec{r}_0 = (-4\hat{i} + 6\hat{j})$ m at $t = 0$, moving with velocity $\vec{v}_0 = (10\hat{i} - 4\hat{j})$ m/s, and has constant acceleration $\vec{a} = (-2\hat{i} + 6\hat{j})$ m/s^2. Find (a) its velocity and position at $t = 4.0$ s, (b) the displacement over those 4 s and its direction, (c) the average velocity, and (d) the instant at which it is moving parallel to the $y$-axis, and its speed then.

Solution:

1. (a) Velocity, from $\vec{v} = \vec{v}_0 + \vec{a}t$, applied to each component: $$v_x = 10 + (-2)(4) = 2\ \text{m/s}, \qquad v_y = -4 + (6)(4) = 20\ \text{m/s}$$ $$\vec{v}(4) = (2\hat{i} + 20\hat{j})\ \text{m/s}, \qquad |\vec{v}| = \sqrt{4 + 400} = 20.1\ \text{m/s}$$ Both components positive, so the direction is $\tan^{-1}(20/2) = 84.3°$ from $+x$ — almost straight up the $y$-axis.
2. Position, from $\vec{r} = \vec{r}_0 + \vec{v}_0 t + \tfrac{1}{2}\vec{a}t^2$: $$x = -4 + (10)(4) + \tfrac{1}{2}(-2)(16) = -4 + 40 - 16 = 20\ \text{m}$$ $$y = 6 + (-4)(4) + \tfrac{1}{2}(6)(16) = 6 - 16 + 48 = 38\ \text{m}$$ $$\vec{r}(4) = (20\hat{i} + 38\hat{j})\ \text{m}$$
3. (b) Displacement is final minus initial position: $$\Delta\vec{r} = (20 - (-4))\hat{i} + (38 - 6)\hat{j} = (24\hat{i} + 32\hat{j})\ \text{m}$$ $$|\Delta\vec{r}| = \sqrt{576 + 1024} = \sqrt{1600} = 40\ \text{m}$$ Both components positive $\Rightarrow$ first quadrant $\Rightarrow$ direction $\tan^{-1}(32/24) = 53.1°$ from $+x$.
4. (c) Average velocity: $$\vec{v}_{avg} = \frac{\Delta\vec{r}}{\Delta t} = \frac{24\hat{i} + 32\hat{j}}{4.0} = (6\hat{i} + 8\hat{j})\ \text{m/s}$$ with magnitude $\sqrt{36+64} = 10$ m/s, also at $53.1°$ — necessarily the same direction as the displacement, since dividing by a positive scalar cannot turn a vector.
5. Notice what the average velocity is not. It is not the velocity at any of the instants we have computed: at $t=0$ the speed was $\sqrt{100+16} = 10.8$ m/s in a quite different direction, and at $t=4$ s it was 20.1 m/s. Average velocity is a displacement-per-time bookkeeping number, not a snapshot.
6. (d) Moving parallel to the $y$-axis means $v_x = 0$: $$10 - 2t = 0 \quad\Rightarrow\quad t = 5.0\ \text{s}$$ At that instant $v_y = -4 + (6)(5.0) = 26$ m/s, so the particle is moving at 26 m/s straight along $+y$.
7. A cross-check on step 4. Since $\vec{a}$ is constant, $\vec{v}_{avg}$ must equal $\tfrac{1}{2}(\vec{v}_0 + \vec{v})$: $$\tfrac{1}{2}\big[(10\hat{i}-4\hat{j}) + (2\hat{i}+20\hat{j})\big] = \tfrac{1}{2}(12\hat{i} + 16\hat{j}) = 6\hat{i} + 8\hat{j}\ \checkmark$$

Final Answer: (a) $\vec{v} = (2\hat{i}+20\hat{j})$ m/s and $\vec{r} = (20\hat{i}+38\hat{j})$ m; (b) displacement 40 m at $53.1°$; (c) average velocity 10 m/s at $53.1°$; (d) at $t = 5.0$ s, moving at 26 m/s.

Takeaway: The vector equations of motion are used exactly as you used the scalar ones in Chapter 2 — once for $x$, once for $y$, never mixing the two. And step 7 is the check worth building into your habits: with constant acceleration only, $\vec{v}_{avg} = \tfrac{1}{2}(\vec{v}_0 + \vec{v})$, and it catches arithmetic slips instantly. (Example 16 has just shown you it is only valid for constant $\vec{a}$.)

Example 19: Reading "speeding up" and "slowing down" off the vectors

A particle moves in the x-y plane with $\vec{r}(t) = 6t\,\hat{i} + (8t - 5t^2)\,\hat{j}$ metres. (a) Find $\vec{v}(t)$ and $\vec{a}$. (b) Find the angle between $\vec{v}$ and $\vec{a}$ at $t = 0$ and at $t = 1.0$ s, and say in each case whether the particle is speeding up or slowing down. (c) At what instant are $\vec{v}$ and $\vec{a}$ perpendicular, and what is the speed then?

Solution:

1. (a) Differentiate once, then again: $$\vec{v} = \frac{d\vec{r}}{dt} = 6\,\hat{i} + (8 - 10t)\,\hat{j}\ \ \text{m/s}, \qquad \vec{a} = \frac{d\vec{v}}{dt} = -10\,\hat{j}\ \ \text{m/s}^2$$ The acceleration is constant and points along $-\hat{j}$ at all times. (This particle is a projectile launched with $\vec{v}_0 = 6\hat{i}+8\hat{j}$ under $g = 10$ m/s^2.)
2. (b) At $t = 0$: $\vec{v} = 6\hat{i} + 8\hat{j}$, so $|\vec{v}| = \sqrt{36+64} = 10$ m/s in the direction $\tan^{-1}(8/6) = 53.1°$. The acceleration points at $-90°$. The angle between them is $$53.1° - (-90°) = 143.1°$$ Obtuse, so $\vec{a}$ has a component opposing $\vec{v}$: the particle is slowing down.
3. At $t = 1.0$ s: $\vec{v} = 6\hat{i} + (8 - 10)\hat{j} = 6\hat{i} - 2\hat{j}$, so $|\vec{v}| = \sqrt{36+4} = 6.32$ m/s in the direction $-18.4°$ (fourth quadrant, since $v_x>0$ and $v_y<0$). The angle to $\vec{a}$ at $-90°$ is $$-18.4° - (-90°) = 71.6°$$ Acute, so the particle is speeding up. Confirm it against the speeds themselves: 10 m/s at $t=0$, then 6.32 m/s at $t=1$ s — so it slowed down first and is now on the way back up.
4. (c) Perpendicular means the angle is exactly $90°$. With $\vec{a}$ pointing straight down, $\vec{v}$ must be exactly horizontal, i.e. $v_y = 0$: $$8 - 10t = 0 \quad\Rightarrow\quad t = 0.80\ \text{s}$$
5. The speed there is whatever is left, namely the horizontal part on its own: $$|\vec{v}| = \sqrt{6^2 + 0^2} = 6.0\ \text{m/s}$$
6. Recognise the instant. $v_y = 0$ with $\vec{a}$ downward is the top of the trajectory, and 6.0 m/s is the minimum speed of the entire motion — the smallest it ever gets, because $v_x$ never changes and $v_y^2$ is at its least possible value, zero. Everything in step 3 falls into place: before $t = 0.80$ s the speed is falling, after it the speed is rising.

Final Answer: (a) $\vec{v} = 6\hat{i} + (8-10t)\hat{j}$ m/s, $\vec{a} = -10\hat{j}$ m/s^2. (b) $143.1°$ at $t=0$ (slowing down); $71.6°$ at $t=1.0$ s (speeding up). (c) At $t = 0.80$ s, speed 6.0 m/s.

Takeaway: The angle between $\vec{v}$ and $\vec{a}$ is a speedometer needle: acute means speeding up, obtuse means slowing down, and exactly $90°$ means the speed is momentarily neither — a turning point of speed, the minimum here. [JEE Tip] That $90°$ condition locates the top of any projectile without ever writing $T/2$, and it is the same condition that lets uniform circular motion run at constant speed forever (Example 32).

Example 20: One throw, every number in it

A ball is projected from level ground with speed $v_0 = 36$ m/s at $\theta_0 = 40°$ above the horizontal. Take $g = 9.8$ m/s^2 and neglect air resistance. Find (a) the components of the initial velocity, (b) the time of flight, (c) the maximum height, (d) the horizontal range, and (e) the velocity 1.0 s after launch.

Solution:

1. (a) Resolve the launch velocity. Always the first line of a projectile solution: $$v_{0x} = v_0\cos\theta_0 = 36\cos 40° = 36(0.7660) = 27.6\ \text{m/s}$$ $$v_{0y} = v_0\sin\theta_0 = 36\sin 40° = 36(0.6428) = 23.1\ \text{m/s}$$ From here on, $v_x$ never changes and only $v_y$ feels gravity.
2. (b) Time of flight. The ball returns to the launch height, so the $y$-displacement over the whole flight is zero: $$T = \frac{2v_0\sin\theta_0}{g} = \frac{2(23.14)}{9.8} = 4.72\ \text{s}$$
3. (c) Maximum height, reached at $T/2 = 2.36$ s when $v_y = 0$: $$H = \frac{v_0^2\sin^2\theta_0}{2g} = \frac{(23.14)^2}{2(9.8)} = \frac{535.5}{19.6} = 27.3\ \text{m}$$
4. (d) Horizontal range — the horizontal motion is uniform, so it is just speed times time: $$R = v_{0x} \times T = 27.58 \times 4.72 = 130\ \text{m}$$ Check against the formula: $R = \dfrac{v_0^2\sin 2\theta_0}{g} = \dfrac{1296\sin 80°}{9.8} = \dfrac{1296(0.9848)}{9.8} = 130$ m. $\checkmark$
5. (e) Velocity at $t = 1.0$ s. The horizontal part is untouched; the vertical part has lost $gt$: $$v_x = 27.6\ \text{m/s}, \qquad v_y = 23.14 - (9.8)(1.0) = 13.3\ \text{m/s}$$ $$|\vec{v}| = \sqrt{27.58^2 + 13.34^2} = \sqrt{760.5 + 178.0} = \sqrt{938.5} = 30.6\ \text{m/s}$$ Direction: $\tan^{-1}(13.34/27.58) = 25.8°$ above the horizontal (both components positive, so it is still rising).

Final Answer: (a) 27.6 m/s and 23.1 m/s; (b) $T = 4.72$ s; (c) $H = 27.3$ m; (d) $R = 130$ m; (e) 30.6 m/s at $25.8°$ above the horizontal.

Takeaway: Every projectile problem starts with the same two lines — resolve $v_0$ into $v_{0x}$ and $v_{0y}$ — and then becomes bookkeeping. Note how the launch angle of $40°$ had already fallen to $25.8°$ after just one second: the direction of motion changes continuously even though the acceleration never does.

Example 21: The complementary pair, from a measured range

A ball thrown at $33°$ to the horizontal lands 90 m away on level ground. Take $g = 9.8$ m/s^2. Find (a) the launch speed, (b) the other launch angle that gives the same 90 m range, (c) the maximum height for each of the two throws, and (d) verify the two neat relations $H_1 + H_2 = \dfrac{v_0^2}{2g}$ and $R = 4\sqrt{H_1 H_2}$.

Solution:

1. (a) Work backwards from the range formula. $$R = \frac{v_0^2\sin 2\theta_0}{g} \quad\Rightarrow\quad v_0^2 = \frac{Rg}{\sin 2\theta_0} = \frac{90 \times 9.8}{\sin 66°} = \frac{882}{0.9135} = 965.5$$ $$v_0 = \sqrt{965.5} = 31.1\ \text{m/s}$$
2. (b) The complementary angle. Range depends on $\sin 2\theta_0$, and $\sin(180° - x) = \sin x$, so $2\theta_0' = 180° - 66° = 114°$, giving $$\theta_0' = 57° \qquad (\text{and indeed}\ 33° + 57° = 90°)$$
3. (c) Maximum heights. With $\dfrac{v_0^2}{2g} = \dfrac{965.5}{19.6} = 49.26$ m, $$H_1 = 49.26\sin^2 33° = 49.26(0.2966) = 14.6\ \text{m}$$ $$H_2 = 49.26\sin^2 57° = 49.26(0.7034) = 34.6\ \text{m}$$ The steeper throw goes 2.37 times higher for the identical range.
4. (d) First relation. Since $\sin^2 33° + \sin^2 57° = \sin^2 33° + \cos^2 33° = 1$, $$H_1 + H_2 = 14.6 + 34.6 = 49.3\ \text{m} = \frac{v_0^2}{2g}\ \checkmark$$ which is the maximum height a vertical throw at the same speed would reach.
5. Second relation: $$4\sqrt{H_1H_2} = 4\sqrt{14.61 \times 34.65} = 4\sqrt{506.3} = 4(22.5) = 90.0\ \text{m} = R\ \checkmark$$
6. A bonus difference. The two flight times are $T_1 = \dfrac{2(31.07)\sin 33°}{9.8} = 3.45$ s and $T_2 = \dfrac{2(31.07)\sin 57°}{9.8} = 5.32$ s. Their ratio is $\dfrac{5.32}{3.45} = 1.54 = \tan 57°$.

Final Answer: (a) 31.1 m/s; (b) $57°$; (c) 14.6 m and 34.6 m; (d) both relations check out, with $H_1 + H_2 = 49.3$ m and $4\sqrt{H_1H_2} = 90.0$ m.

Takeaway: Equal ranges do not mean equal flights. The complementary pair share only $R$; the steeper one goes much higher and stays up $\tan\theta_2$ times longer. [JEE Tip] Remember $H_1 + H_2 = \dfrac{v_0^2}{2g}$ and $R = 4\sqrt{H_1H_2}$ — either one lets you jump straight from two heights to a speed or a range with no angle in sight.