A man whose mass is $m k g$ jumps vertically into air from the sitting position in which his centre of mass is at a height $h_{1}$ from the ground. When his feet are just about to level the ground his center of mass is at height $h_{2}$ from the ground and finally center of mass rises to $h_{3}$ above the ground when he is at the top of the jump. What is the average upward force exerted by the ground on him?

\frac{m g (h_{3} - h_{1})}{(h_{3} - h_{2})}

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\frac{m g (h_{3} - h_{1})}{h_{3}}

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\frac{m g (h_{3} - h_{1})}{(h_{2} - h_{1})}

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Solution

The correct option is C $\frac{m g (h_{3} - h_{1})}{(h_{2} - h_{1})}$
From $h_{1}$ to $h_{2}$ , let the average force be $F$ . Acceleration $a$ will be upward.
$F - m g = m a \Rightarrow a = \frac{F - m g}{m}$
Let $v$ be the velocityof the centre of mass when the contact is left:
$v^{2} = 2 a (h_{2} - h_{1})$
$\Rightarrow v^{2} = 2 (\frac{F - m g}{m}) (h_{2} - h_{1})$
Now he is free fall. Further the height raised $= h_{3} - h_{2}$
So $h_{3} - h_{2} = \frac{v^{2}}{2 g}$
$\Rightarrow h_{1} - h_{2} = (\frac{F - m g}{m g}) (h_{2} - h_{1})$
$\Rightarrow F = \frac{m g (h_{3} - h_{1})}{(h_{2} - h_{1})}$

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