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

Answer:

v = 0.363 m/s

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,

$h=ut+\dfrac{1}{2}at^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2\times 0.55}{9.8}} \\\\t=0.33\ s$

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

$v_x=\dfrac{x}{t}\\\\v_x=\dfrac{0.12}{0.33}\\\\v_x=0.363\ m/s$

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