Respuesta :
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,
[tex]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[/tex]
It can be assumed to find the initial horizontal velocity of the tennis ball. It can be given by :
[tex]v_x=\dfrac{x}{t}\\\\v_x=\dfrac{0.12}{0.33}\\\\v_x=0.363\ m/s[/tex]
Hence, the initial horizontal velocity is 0.363 m/s.
