$1.4.11.$ The body is dropped over the slab at a height $h$ from it. The slab moves vertically upwards at a speed of $u$. Determine the time between two consecutive impacts of the body on the slab. The blows are absolutely elastic.
Let's move to the reference frame associated with the plate. At the initial moment of time, the relative velocity is $u$ and is directed downwards.
From the law of conservation of energy, we find the velocity of the body $v$ at the moment before the impact
$$ \frac{mv^2}{2} = \frac{mu^2}{2} + mgh $$
Where is the desired velocity
$$v = \sqrt{u^2+2gh}$$
In this inertial reference frame, after an elastic rebound, the relative velocity will only change direction.
A body thrown vertically will land after time
$$ t = \frac{2v}{g} $$
Time sought:
$$ \fbox{$t = \frac{2\sqrt{u^2 + 2gh}}{g}$} $$
$$t = 2\sqrt{u^2/g^2 + 2h/g}$$