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.

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 cosθ

                                   = 24.5 * cos 22.5°

                                   = 24.5 * 0.924

                                   = 22.64 m/s

Vertical velocity, = u sinθ

                                 = 24.5 * sin 22.5°

                                 = 24.5 * 0.383

                                 = 9.38 m/s

We know,

To calculate the height:

Therefore, height of the ball would be 4.32 m

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