91Ó°ÊÓ

It is given that,

1. The mass of the man$M=80\mathrm{kg}$
2. Height of the window from patio$h=0.5m$

The distance covered during landing to come to the stoprole="math" localid="1657023846178" $∆y=0.02\mathrm{m}$

This problem is based on the kinematic equations of motion that describe the motion under constant acceleration. Using these equations of motion, the acceleration and force can be computed.

Formulae:

The final velocity in kinematic equation can be written as

$v_f^2={v_0}^2+2ah$ …(¾±)

where, $v_f$is the final velocity, $v_0$is the initial velocity, $a$is an acceleration and$∆y$is displacement.

During the free fall, the time taken by the man can be calculated as follows:

The equation (i),

$v_f^2={v_0}^2+2ah$

Initial velocity

$v_0=0$

$$\begin{gathered} {v_f}^2=2×9.8\mathrm{m}/{\mathrm{s}}^2×0.5\mathrm{m} \\ =9.8{\mathrm{m}}^2/{\mathrm{s}}^2 \\ {\mathrm{v}}_{\mathrm{f}}=3.13\mathrm{m}/\mathrm{s} \end{gathered}$$

The average acceleration of the man when his feet first touch the patio to when he stops is calculated using equation (i),

$$\begin{gathered} v_f^2{=}_0^2+2a(∆y] \\ 0={(3.13\mathrm{m}/\mathrm{s})}^2-2\mathrm{a}(0.02\mathrm{m}) \\ \mathrm{a}=\frac{{(3.13\mathrm{m}/\mathrm{s})}^2}{0.04\mathrm{m}} \\ \mathrm{a}=244.9 \\ ~245\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the average acceleration of the man when his feet first touch the patio to when he stops is$245\mathrm{m}/{\mathrm{s}}^2$.

The magnitude of the average stopping force exerted on him by the patio is,

$$\begin{gathered} F-Mg=Ma \\ F=Ma+Mg \\ F=80\mathrm{kg}·(245+9.8)\mathrm{m}/{\mathrm{s}}^2 \\ F=20.4\mathrm{kN} \end{gathered}$$

Hence, the magnitude of the average stopping force exerted on him by the patio is$20.4\mathrm{kN}$.