A ball of mass $= 100 g m$ is released from a height $h_{1} = 2.5 m$ above the ground level and then rebounds to a height $h_{2} = 0.625 m$ . The time of contact of the ball with the ground is $Δ t = 0.01 s e c$ . Find the impulsive (impact) force offered by the ball on the ground.

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Solution

$\to F = \frac{m ({\to v}_{2} - {\to v}_{1})}{δ t}$
where $∣ ∣ {\to v}_{2} - {\to v}_{1} ∣ ∣ = v_{2} + v_{1}$ where $v_{1} = \sqrt{2 g h_{1}}; v_{2} = \sqrt{2 g h_{2}}$
$\Rightarrow F = \frac{m (\sqrt{2 g h_{1}} + \sqrt{2 g h_{2}})}{Δ t} = \frac{1}{10} [\frac{\sqrt{2 (8.8) (2.5)} + \sqrt{2 (8.8) (0.625)}}{0.01}] = 105 N$

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A ball of mass =100gm is released from a height h1=2.5m above the ground level and then rebounds to a height h2=0.625m. The time of contact of the ball with the ground is Δt=0.01sec. Find the impulsive (impact) force offered by the ball on the ground.

→F=m(→v2−→v1)δtwhere ∣∣→v2−→v1∣∣=v2+v1 where v1=√2gh1;v2=√2gh2⇒F=m(√2gh1+√2gh2)Δt=110[√2(8.8)(2.5)+√2(8.8)(0.625)0.01]=105N

A ball of mass $= 100 g m$ is released from a height $h_{1} = 2.5 m$ above the ground level and then rebounds to a height $h_{2} = 0.625 m$ . The time of contact of the ball with the ground is $Δ t = 0.01 s e c$ . Find the impulsive (impact) force offered by the ball on the ground.