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 38.5 ∘ from the ground with an initial speed of 21.5 m / s. At what height above its initial position does the tennis ball hit the wall? Ignore any effects of air resistance.  

Answer 1

Let us consider the upwards direction as positive and the downwards direction as negative.

Gravitational acceleration = g = -9.81 m/s²

Initial velocity of the tennis ball = V = 21.5 m/s

Angle the tennis ball is thrown at = = 38.5^o

Initial horizontal velocity of the tennis ball = V_x = VCos = (21.5)Cos(38.5) = 16.826 m/s

Initial vertical velocity of the tennis ball = V_y = VSin = (21.5)Sin(38.5) = 13.384 m/s

Horizontal distance of you from the wall = R = 17.5 m

Time taken by the ball to reach the wall = T

There is no horizontal force acting on the tennis ball therefore the horizontal velocity of the tennis ball remains constant.

R = V_xT

17.5 = (16.826)T

T = 1.04 sec

Height the tennis ball hits the wall at = H

H = V_yT + gT²/2

H = (13.384)(1.04) + (-9.81)(1.04)²/2

H = 8.61 m

Height at which the tennis ball hits the wall = 8.61 m