If a man of mass M jumps to the ground from a height h and his centre of mass moves a distance of x during the impact, the average force acting on him (assuming his retardation to be constant during his impact with the ground) is

$\frac{M g h}{x}$

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$\frac{M g x}{h}$

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$M g {(\frac{h}{x})}^{2}$

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$M g {(\frac{x}{h})}^{2}$

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Solution

The correct option is A
$\frac{M g h}{x}$

Since the centre of mass moves through a distance x, the average force $F$ is given by $F x = M g h$
$\Rightarrow F = \frac{M g h}{x}$
Hence, the correct choice is (a).

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If a man of mass M jumps to the ground from a height h and his centre of mass moves a distance of x during the impact, the average force acting on him (assuming his retardation to be constant during his impact with the ground) is

The correct option is A Mghx Since the centre of mass moves through a distance x, the average force F is given by Fx=Mgh ⇒F=Mghx Hence, the correct choice is (a).

The correct option is A
$\frac{M g h}{x}$

Since the centre of mass moves through a distance x, the average force $F$ is given by $F x = M g h$
$\Rightarrow F = \frac{M g h}{x}$
Hence, the correct choice is (a).