Question

A tennis ball is struck and departs from the racket horizontally with a speed of 28.9 m/s. The ball hits the court at a horizontal distance of 18.8 m from the racket. How far above the court is the tennis ball when it leaves the racket?

Answer 1

Let us consider the downward direction as positive.

Gravitational acceleration = g = 9.81 m/s²

Initial velocity of the tennis ball = V₀ = 28.9 m/s

The ball moves horizontally initially.

Initial horizontal velocity of the tennis ball = V_x0 = 28.9 m/s

Initial vertical velocity of the tennis ball = V_y0 = 0 m/s

Horizontal distance covered by the tennis ball = R = 18.8 m

Initial height of the tennis ball = H

Time taken by the ball to hit the court.

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

R = V_x0T

18.8 = (28.9)T

T = 0.65 sec

H = V_y0T + gT²/2

H = (0)(0.65) + (9.81)(0.65)²/2

H = 2.07 m

Height of the tennis ball above the court when it leaves the racket = 2.07 m