Question

A man whose mass is $m$ kg jumps vertically into air from a sitting position in which his centre of mass is at a height $h_1$ from the ground. When his feet are just about to leave the ground his centre of mass is $h_2$ from the ground and finally rises to $h_3$ when he is at the top of the jump. What is the average upward force exerted by the ground on him?

Answer 1

Let velocity of mass at height $h_2$ is $\left(V\right),$ at height $h_3$ velocity is zero.

$\Rightarrow v_0^2-v^2=2as$

$0-v^2=2(-y)(h_3-h_2)$

$v^2=2g(h_3-h_2)$

$\Rightarrow v=\sqrt{2g(h_3-h_2)}$

Let average fore is a from $h_1$ to $h_2$

Change in K.E $=$ work done by force.

$\Rightarrow (f-mg)\times (h_2-h)=\frac{1}{2}m{v}^2$

$(f-mg)(h_2-h_1)=\frac{1}{2}m/2g(h_3-h_2)$

$f-mg=mg\frac{(h_3-h_2)}{(h_2-h_1)}$

$\Rightarrow f=mg+mg\frac{(h_3-h_2)}{(h_2-h_1)}$

$=mg(1+\frac{h_3-h_2}{h_2-h_1})$

$=mg\left(\frac{h_3-h_1}{h_2-h_1}\right).$

Hence, solved.