Question

A tennis ball rolls off the edge of a table. the table is 0.55m tall and the tennis ball lands 0.12m away from the table.

Answer 1

v = 0.363 m/s

Step-by-step explanation:

Given that,

The table is 0.55m tall and the tennis ball lands 0.12m away from the table.

Here, u = 0 (at rest) for initial vertical velocity as it rolls off the edge of a table.

Let t is the time to fall from the vertical height. So,

`h=ut+(1)/(2)at^2\\\\t=\sqrt{(2h)/(g)} \\\\t=\sqrt{(2* 0.55)/(9.8)} \\\\t=0.33\ s`

It can be assumed to find the initial horizontal velocity of the tennis ball. It can be given by :

`v_x=(x)/(t)\\\\v_x=(0.12)/(0.33)\\\\v_x=0.363\ m/s`

Hence, the initial horizontal velocity is 0.363 m/s.