Question

A 80 kgman drops to a concrete patio from a window 0.50 mabove the patio. He neglects to bend his knees on landing, taking 2.0 cmto stop. (a) What is his average acceleration from when his feet first touch the patio to when he stops? (b) What is the magnitude of the average stopping force exerted on him by the patio?

Short answer

1. The average acceleration of the man when his feet first touch the patio to when he stops is$245\mathrm{m}/{\mathrm{s}}^2$
2. The magnitude of the average stopping force exerted on him by the patio is$20.4k\mathrm{N}$

Step-by-step solution

Given information

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" $∆y=0.02\mathrm{m}$

Determining the concept

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

$v_f^2={v_0}^2+2ah$ …(i)

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

(a) Determining the average acceleration of the man when his feet first touch the patio to when he stops

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

$v_0=0$

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

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$245\mathrm{m}/{\mathrm{s}}^2$.

(b) Determining the magnitude of the average stopping force exerted on him by the patio

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$20.4\mathrm{kN}$.