Chapter 5, Problem 55QAP

To determine

(a)

The terminal velocity of the person with parachute.

Answer to Problem 55QAP

The terminal velocity of the person with parachute is $6.17\text{m}{\text{s}}^{\text{-1}}$

Explanation of Solution

Given:

Mass of the person

- $70.0\text{kg}$

Proportionality constant value

- $12.0{\text{ ms}}^{-1}$

Formula used:

Newton's second law - $F=ma$ ( Used considering the vertical axis)

  $F_{\text{drag on person }}+\left(-W_{\text{person}}\right)=0$

  $F_{\text{drag}}=c{v}^2$

Calculation:

  $\begin{array}{l}{F}_{\text{drag on person }}+\left(-W_{\text{person}}\right)=0\\ \text{For terminal velocity}\\ c{v}^2{}_{\text{terminal}}-\left(mg\right)=0\end{array}$

To increase the velocity to $20.0{\text{ ms}}^{-1}.$

  $\begin{array}{l}{F}_{\text{drag on person }}+\left(-W_{\text{person}}\right)=0\\ \text{For terminal velocity}\\ F_{\text{drag on person }}+\left(-mg\right)=0\\ c{v}^2{}_{\text{terminal}}+\left(\left(-50.0\text{kg}\right)\cdot \left(9.80{\text{ ms}}^{-2}\right)\right)=0\\ c{v}^2{}_{\text{terminal}}=686{\text{ kgms}}^{-2}\\ v^2{}_{\text{terminal}}=\frac{686{\text{ kgms}}^{-2}}{c}\\ v_{\text{terminal}}=\sqrt{\frac{686{\text{ kgms}}^{-2}}{c{\text{ kgm}}^{-1}}}\\ v_{\text{terminal}}=\frac{1}{\sqrt{c}}\cdot \left(26.192\right){\text{ms}}^{-1}\\ v_{\text{terminal}}=\frac{1}{\sqrt{18}}\cdot \left(26.192\right){\text{ms}}^{-1}=6.17\text{m}{\text{s}}^{\text{-1}}\end{array}$

Conclusion:

Thus, the terminal velocity of the person with parachute is $v_{\text{terminal}}=6.17\text{m}{\text{s}}^{\text{-1}}$

To determine

(b)

The value of the proportionality constant c when the terminal velocity is $50.0{\text{ ms}}^{-1}.$

Answer to Problem 55QAP

The value of the proportionality constant c when the terminal velocity is $50.0{\text{ ms}}^{-1}$ is $0.2744\text{}\text{ kg}{\text{ms}}^{-1}$

Explanation of Solution

Given:

Mass of the person

- $70.0\text{kg}$

Terminal velocity

- $50.0{\text{ ms}}^{-1}$

Formula used:

Newton's second law - $F=ma$ ( Used considering the vertical force components)

  $F_{\text{drag on person }}+\left(-W_{person}\right)=0$

  $F_{\text{drag}}=c{v}^2$

Calculation:

  $\begin{array}{l}{F}_{\text{drag on person }}+\left(-W_{person}\right)=0\\ \text{For terminal velocity}\\ c{v}^2{}_{ter\min al}-mg=0\end{array}$

  $\begin{array}{l}c\cdot \left({\left(50{\text{ ms}}^{-1}\right)}^2\right)+\left(\left(-70.0\text{kg}\right)\cdot \left(9.80{\text{ ms}}^{-2}\right)\right)=0\\ c=\frac{686{\text{ kgms}}^{-2}}{2500{\text{ m}}^2{\text{s}}^{-2}}\\ c=0.2744\text{}\text{ kg}{\text{ms}}^{-1}\end{array}$

Conclusion:

Thus, the value of the proportionality constant c when the terminal velocity is 50.0 m/s is $0.2744\text{}\text{ kg}{\text{ms}}^{-1}$