Question

A 1 Kg steel ball 10m above the floor is released, falls,
strikes the floor, and rises to a maximum height of 2.5m
Momentum transfered from the ball to the floor in the
collision.​

Answer 1

Answer:

I hope it is helpful for you dera

Answer 2

Answer:

The velocity with which it rebounds is given by $2100\ m / \mathrm{s}^{2}$

Explanation:

Let u be the velocity with which the ball hits the ground, then

$\begin{aligned}&u^{2}=\mathbf{2 g h} \\&=2 \times 9.8 \times 10=196 \\&\therefore \quad u=14 \mathrm{~m} / \mathrm{sec}\end{aligned}$

If $v$ be the velocity with which it rebounds, then

$\begin{aligned}&\mathbf{v}^{2}=2 \times 9.8 \times 2.5=49 \\&\Rightarrow v=7 \mathrm{~m} / \mathrm{sec} \\&\therefore \quad \Delta \mathrm{v}=(\mathrm{v}-\mathrm{u}) \\&=(7 \mathrm{~m} / \mathrm{sec})-(-14 \mathrm{~m} / \mathrm{sec}) \\&=21 \mathrm{~m} / \mathrm{sec} \\&\therefore \quad \mathrm{a}=\frac{\Delta \mathrm{v}}{\Delta \mathrm{t}}=\frac{21}{0.01} \\&=\mathbf{2 1 0 0 m} / \mathrm{s}^{2}\end{aligned}$