Question

A cricketer of height 2.5 m throws a ball at an angle of 30° with the horizontal such that it is received by another cricketer of the same height standing at a distance of 50 m from the first one. The maximum height attained by the ball is (tan 30° = 0.577).

• A. 7.9 m
• B. 10 m
• C. 10.7 m
• D. 9.7 m

Hint:

In projectile motion, the horizontal and vertical motions are independent of each other. The maximum height is achieved when the vertical velocity component is at its peak, and the time of flight can be found using the range equation.

Answer 1

The Correct Option is D

We are given the following data:
- Angle of projection, \( \theta = 30^\circ \) - Initial height, \( h_0 = 2.5 \, \text{m} \) - Horizontal distance, \( R = 50 \, \text{m} \) - \( \tan 30^\circ = 0.577 \) - Gravitational acceleration, \( g = 9.8 \, \text{m/s}^2 \)
1. Find the time of flight \( t \): The horizontal velocity component \( v_x = v \cos(\theta) \), where \( v \) is the initial velocity. The horizontal distance is related to the time of flight by: \[ R = v_x \times t \] \[ 50 = v \cos(30^\circ) \times t \] \[ 50 = v \times 0.866 \times t \]
Thus, \[ t = \frac{50}{v \times 0.866} \]
2. Vertical motion: For vertical motion, we use the kinematic equation for maximum height \( h_{\text{max}} \): \[ h_{\text{max}} = h_0 + \frac{v_y^2}{2g} \] where \( v_y = v \sin(\theta) \). Substituting \( v_y = v \sin(30^\circ) = 0.5v \), the maximum height equation becomes: \[ h_{\text{max}} = 2.5 + \frac{(0.5v)^2}{2g} = 2.5 + \frac{0.25v^2}{2 \times 9.8} \]
3. Use the range equation to find \( v \): The range equation for projectile motion is: \[ R = \frac{v^2 \sin(2\theta)}{g} \] Substituting \( R = 50 \), \( \theta = 30^\circ \), and \( g = 9.8 \, \text{m/s}^2 \): \[ 50 = \frac{v^2 \sin(60^\circ)}{9.8} \] \[ 50 = \frac{v^2 \times 0.866}{9.8} \] \[ v^2 = \frac{50 \times 9.8}{0.866} = 566.65 \] \[ v = 23.8 \, \text{m/s} \]
4. Now, calculate the maximum height attained: Substitute \( v = 23.8 \, \text{m/s} \) into the equation for maximum height: \[ h_{\text{max}} = 2.5 + \frac{0.25 \times (23.8)^2}{2 \times 9.8} \] \[ h_{\text{max}} = 2.5 + \frac{0.25 \times 566.44}{19.6} \] \[ h_{\text{max}} = 2.5 + \frac{141.61}{19.6} = 2.5 + 7.2 = 9.7 \, \text{m} \]
Thus, the maximum height attained by the ball is \( 9.7 \, \text{m} \).