Chapter 7, Problem 21P

To determine

The maximum height can a person jump without breaking the lower leg bone of either leg.

Answer to Problem 21P

The maximum height would be $h_{\max }=69\text{m}$ .

Explanation of Solution

Given:

Weight of a person is $m=75\text{kg}$ . Breaking distance from standing to the seated position is $h_{stop}=0.60\text{m}$ . Breaking strength is $170\times {10}^6{\text{N/m}}^{\text{2}}$ . The smallest cross-sectional area is $2.5\times {10}^{-4}{\text{m}}^{\text{2}}$ .

Formula used:

According to the law of conservation of energy,

Kinetic Energy = Potential Energy

  $\frac{1}{2}m{v}_{final}{}^2=mg{h}_{\max }$

The second equation of motion is

  $s=h_{stop}=ut+\frac{1}{2}a{t}^2$

Calculation:

Using the law of conservation of energy,

  $\begin{array}{l}\frac{1}{2}m{v}_{final}{}^2=mg{h}_{\max }\\ v_{final}=\sqrt{2g{h}_{\max }}(1)\end{array}$

When a person hits the ground equal and opposite reaction will be applied by the ground. Thus, negative force will act upon the person i.e. in upward direction.

  $\begin{array}{l}F\Delta t=m(v_{initial}-v_{final})\\ F\Delta t=m(0-v_{final})\\ \Delta t=-\frac{m{v}_{final}}{F}(2)\end{array}$

Using Newton’s second law of motion

  $\begin{array}{l}F=ma\\ a=\frac{F}{m}(3)\end{array}$

Now, using second equation of motion to calculate maximum height.

  $h_{stop}=ut+\frac{1}{2}a{t}^2$

Inserting all the values from equations (1), (2), (3).

  $\begin{array}{l}{h}_{stop}=ut+\frac{1}{2}a{t}^2\\ h_{stop}=v_{final}\left(-\frac{m{v}_{final}}{F}\right)+\frac{1}{2}\left(\frac{F}{m}\right){\left(-\frac{m{v}_{final}}{F}\right)}^2\\ h_{stop}=-\frac{2mg{h}_{\max }}{F}+\frac{1}{2}\frac{2mg{h}_{\max }}{F}=-\frac{mg{h}_{\max }}{F}\\ h_{\max }=-\frac{F{h}_{stop}}{mg}\end{array}$

Since downward force is acting in upward direction. It will cancel out the negative sign. Also, force will be distributed in the two legs of the person equally. Thus,

  $\begin{array}{l}{h}_{\max }=\frac{2F{h}_{stop}}{mg}\\ h_{\max }=\frac{2\times \left(170\times {10}^6\right)\times \left(2.5\times {10}^{-4}\right)\times 0.6}{75\times 9.8}\\ h_{\max }=69\text{m}\end{array}$

Conclusion:

Thus, maximum height from which person should fall without breaking any bone is $69\text{m}$ .