Answer:[tex]t=2\frac{-u+\sqrt{u^2+2gh}}{g}[/tex]
Explanation:
Given
Ball is dropped from a height h and plate is moving upward with velocity u
considering plate at rest by providing a velocity in opposite direction thus ball is now moving with a velocity of u.
As acceleration is acting downwards therefore ball will accelerate with g
and time taken to collide with plate is
[tex]h=ut+\frac{gt^2}{2}[/tex]
thus [tex]t=\frac{-2u\pm \sqrt{4u^2+4\times g\times 2h}}{2g}[/tex]
Taking Positive value only
[tex]t=\frac{-2u+\sqrt{4u^2+4\times g\times 2h}}{2g}[/tex]
ball collide and bounces back to original height and then again collides with plate taking time=2t
Thus time between two collisions is equal to
[tex]t=2\frac{-2u+\sqrt{4u^2+4\times g\times 2h}}{2g}[/tex]
[tex]t=2\frac{-u+\sqrt{u^2+2gh}}{g}[/tex]
