Question

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Answer 1

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Answer 2

Answer: The average acceleration will be $a = 500\sqrt{2}\ m/s^2$ in upwards direction.

Explanation:

Given that,

Height h = 10 m

Height after rebound h' = 2.5 m

Time t = 0.01 s

Using equation of motion

$v^2=u^2+2gh$

$v^2=0+2gh$

$v= \sqrt{2gh}$

$v = \sqrt{2\times9.8\times10}$

After rebound,

Initial velocity u= v'

Again using equation of motion

$v^2=u^2-2gh$

$0 = v'^2-2gh'$

$v' = \sqrt{2gh'}$

$v' = \sqrt{2\times9.8\times2.5}$

Average acceleration is

$a = \dfrac{v-v'}{t}$

$a = \dfrac{\sqrt{2\times9.8\times10}-\sqrt{2\times9.8\times2.5}}{0.01}$

$a = 500\sqrt{2}\ m/s^2$

Hence, The average acceleration will be $a = 500\sqrt{2}\ m/s^2$ in upwards direction.