Question

You stand 17.5 m from a wall holding a tennis ball. You throw the tennis ball at the wall at an angle of 22.5∘ from the ground with an initial speed of 24.5 m/s. At what height above its initial position does the tennis ball hit the wall? Ignore any effects of air resistance.

Answer 1

The height of the ball would be 4.32 m

Explanation:

Given-

Distance from the ball, s = 17.5 m

Angle of projection, θ = 22.5°

Initial speed, u = 24.5 m/s

Height, h = ?

Let t be the time taken.

Horizontal speed, $u_{x}$ = u cosθ

                                   = 24.5 * cos 22.5°

                                   = 24.5 * 0.924

                                   = 22.64 m/s

Vertical velocity, $u_{y}$ = u sinθ

                                 = 24.5 * sin 22.5°

                                 = 24.5 * 0.383

                                 = 9.38 m/s

We know,

$x = u * cos (theta) * t$

$17.5 = 22.64 * t\\\\t = 0.77s$

To calculate the height:

$h = ut - \frac{1}{2}gt^2$

$h = u sin (theta)t - \frac{1}{2} gt^2$

$h = 9.38 * 0.77 - \frac{1}{2} * 9.8 * (0.77)^2\\ \\h = 7.22 - 2.90\\\\h = 4.32m$

Therefore, height of the ball would be 4.32 m