Question

A tennis ball is struck in such a way that it leaves the racket with a speed of 31.0m/s in the horizontal direction. When the ball hits the court, it is a distance of 13.3m from the racket. Find the height of the racket ball when it left the racket.

Answer 1

Height of the racket ball = 0.86 m

Step-by-step explanation:

Given:

Speed of the tennis ball,
`v_x`= 31 m/s

Distance covered,
`R_x` = 13.3 m

We have to find the height of the racket ball when it left the racket.

Lets say that the time taken by the ball to hit the court be 't' seconds.

⇒
`time=(distance)/(speed)`

⇒
`time=(R_x)/(v_x)`

⇒
`t=(13.3\ m)/(31.0\ ms^-^1)`

⇒
`t=0.42` seconds

Now we have to find the height lets say that the height is 'h' meter.

⇒
`h_y=u_yt + (a_yt^2)/(2)` ...second equation of motion

⇒
`h_y=0 + (gt^2)/(2)` ...initial velocity = 0 and acceleration = gravity

⇒
`h_y=(9.8(0.42)^2)/(2)`

⇒
`h_y=(1.72)/(2)`

⇒
`h_y=0.86` meter.

So the height of the racket ball when it left the racket is of 0.86 m.