A ball is projected from the ground with a speed of 25 m/s. Two seconds later it just clears a wall of height 5 m. Find a) the angle of projection b) the maximum height attained c) How far beyond the wall, the ball again hits the ground?

Solution:

Given,

Initial velocity of projection $\text{(u) = 25 m/s}$

In time $\text{t = 2 seconds}$, the ball clears a wall of height $\text{h = 5 m}$

Let us consider the acceleration due to gravity (g) be 10 m/s² and $\theta$ be the angle of projection of the ball.

We know,

(a) For angle of projection ($\theta$) of the ball:

$\text{h = u sin}\theta t \ – \ \dfrac{1}{2} \text{gt}^2$

$or, 5 = 25 \sin \theta . 2- \dfrac{1}{2} . 10 . 2^2$

$or, 1 = 10 \sin \theta – 2 . 2$

$or, 1 + 4 = 10 \sin \theta$

$or, 5 = 10 \sin \theta$

$or, \sin \theta = \dfrac{1}{2}$

$or, \sin \theta = \sin 60^o$

$\therefore \theta = 60^o$

Hence, the required angle of projection of the ball was $60^o$.

b) For maximum height attained by the ball:

Let us consider the maximum height attained by the ball be h_max meters. At maximum height, the velocity of the ball is 0 m/s. Therefore, $0 = u – gt$

$\text{or, } t = \dfrac{u}{g}$

$\text{or, } t = \dfrac{25}{10}$

$\therefore t = 2.5 s$

And, $h_\text{max} = \dfrac{1}{2} gt^2$

$= \dfrac{1}{2} . 10 . (2.5)^2$

$= 31.25 m$

Hence, the maximum height attained by the ball was 31.25 meters.

c) Time taken by the fall to hit the ground after 2 seconds of its fall:

We know, Total time of flight (T) = 2t

$\therefore T = 2 (2.5) = 5s$

Hence, the required time taken by the ball to reach the ground beyond the wall is (5 – 2 = 3) seconds.