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

Answer:

Height of the racket ball = 0.86 m

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.

⇒ $time=\frac{distance}{speed}$

⇒ $time=\frac{R_x}{v_x}$

⇒ $t=\frac{13.3\ m}{31.0\ ms^-^1}$

⇒ $t=0.42$ seconds

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

⇒ $h_y=u_yt + \frac{a_yt^2}{2}$     ...second equation of motion

⇒ $h_y=0 + \frac{gt^2}{2}$          ...initial velocity = 0 and acceleration = gravity

⇒ $h_y=\frac{9.8(0.42)^2}{2}$

⇒ $h_y=\frac{1.72}{2}$

⇒ $h_y=0.86$ meter.

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