Projectile Motion

What is Projectile Motion?

A projectile is an object upon which the only force acting is gravity. It is an object thrown into space, either horizontally or at an angle, that moves along a curved path under the action of gravity. We typically neglect air resistance in these calculations.

Analysis of Motion

The key to analyzing projectile motion is to treat the motion in the horizontal (x) and vertical (y) directions independently.

• Horizontal Motion: There is no acceleration ($a_x = 0$). The object moves with a constant horizontal velocity. The equations are: $$v_x = v_{0x} = v_0 \cos(\theta_0)$$ $$x = x_0 + (v_0 \cos\theta_0)t$$
• Vertical Motion: There is a constant downward acceleration due to gravity ($a_y = -g$). The motion is identical to that of an object thrown vertically. $$v_y = v_{0y} - gt = v_0 \sin(\theta_0) - gt$$ $$y = y_0 + (v_0 \sin\theta_0)t - \frac{1}{2}gt^2$$ $$v_y^2 = (v_0 \sin\theta_0)^2 - 2g(y - y_0)$$

Key Formulas (for projectile launched from ground, y₀=0, landing at y=0):

• Equation of Trajectory: By eliminating time 't' from the x and y equations, we get the path of the projectile, which is a parabola. $$y = (\tan\theta_0)x - \frac{g}{2(v_0\cos\theta_0)^2}x^2$$
• Time of Flight ($T_f$): The total time the projectile is in the air. This is twice the time it takes to reach the maximum height. $$T_f = \frac{2v_0\sin\theta_0}{g}$$
• Maximum Height ($H_{max}$): The highest point reached. This occurs when the vertical velocity $v_y = 0$. $$H_{max} = \frac{(v_0\sin\theta_0)^2}{2g}$$
• Horizontal Range (R): The horizontal distance covered during the time of flight. $$R = (v_0 \cos\theta_0) T_f = \frac{v_0^2\sin(2\theta_0)}{g}$$ The range is maximum for a given initial speed $v_0$ when the launch angle $\theta_0 = 45^\circ$.

Example:

A cricket ball is thrown at a speed of $28\ m/s$ in a direction $30^\circ$ above the horizontal. Calculate: (a) the maximum height (b) the time taken by the ball to return to the same level (c) the horizontal range (Take $g = 9.8\ m/s^2$)

Solution: Initial velocity $v_0 = 28\ m/s$, launch angle $\theta_0 = 30^\circ$.

(a) Maximum Height: $H_{max} = \frac{(v_0\sin\theta_0)^2}{2g} = \frac{(28 \times \sin 30^\circ)^2}{2 \times 9.8}$ $H_{max} = \frac{(28 \times 0.5)^2}{19.6} = \frac{14^2}{19.6} = \frac{196}{19.6} = 10\ m$.

(b) Time of Flight: $T_f = \frac{2v_0\sin\theta_0}{g} = \frac{2 \times 28 \times \sin 30^\circ}{9.8} = \frac{2 \times 28 \times 0.5}{9.8} = \frac{28}{9.8} \approx 2.86\ s$.

(c) Horizontal Range: $R = \frac{v_0^2\sin(2\theta_0)}{g} = \frac{28^2 \times \sin(2 \times 30^\circ)}{9.8} = \frac{784 \times \sin 60^\circ}{9.8}$ $R = \frac{784 \times (\sqrt{3}/2)}{9.8} = \frac{392\sqrt{3}}{9.8} \approx 69.3\ m$.