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?

The tennis ball is "2.401 m" far.

According to the question,

Horizontal distance = 19.6 m

Horizontal speed = 28

The time taken by the ball be:

→ t =

By substituting the values,

    =

    = second

By using equation of motion, we get

→ s = ut + gt²

By substituting the values,

     =

     = m

Thus the above answer is right.

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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

i think it would be 5