Question

As a tennis ball is struck, it departs from the racket horizontally with a speed of 28.0 m/s. The ball hits the court at a horizontal distance of 19.6 m from the racket. How far above the court is the tennis ball when it leaves the racket?

Answer 1

solution:

long to travel 19.6m at 28m/s

t =19.6/28

=time taken to drop from rest

s=ut+(1/2)at^2

u=0

a=~10

so s=(1/2)*10*(19.6/28)^2 =2.45m