First slide

Motion of centre of mass

Question

If a man of mass M jumps to the ground from a height ‘h’ and his centre of mass moves a distance ‘x’ after his feet strikes the ground. The average force acting on him is

Moderate

A

$Mg (1 + \frac{h}{x})$
B

$\frac{Mgx}{h}$
C

$Mg {(\frac{h}{x})}^{2}$
D

$Mg {(\frac{x}{h})}^{2}$

Solution

$Final velocity V = \sqrt{2 gh} If ‘ a ’ be the retardation of the man, during the period his feet are in contact with the ground, then$
$0 = V^{2} - 2 ax \Rightarrow a = V^{2} / 2 x = 2 gh / 2 x = gh / x$
$N - Mg = Ma = Mg (\frac{h}{x}) ∴ N = Mg (1 + \frac{h}{x})$

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