• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

Suggested languages for you:

Americas

Europe

Q36P

Expert-verified
Found in: Page 143

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# The terminal speed of a sky diver is ${}^{\mathbf{160}\mathbf{}\mathbf{km}\mathbf{/}\mathbf{h}}$ in the spread-eagle position and ${}^{\mathbf{310}\mathbf{}\mathbf{km}\mathbf{/}\mathbf{h}}$ in the nosedive position. Assuming that the diver’s drag coefficient C does not change from one position to the other, find the ratio of the effective cross-sectional area A in the slower position to that in the faster position.

The ratio of the effective cross-sectional area A in the slower position to that in the faster position is ${}^{3.75}$

See the step by step solution

## Step 1: Given

$a=\frac{2mg}{Cp{{v}^{2}}_{t}}$

which illustrates the inverse proportionality between the area and the speed-squared

thus

${A}_{slow}=310\mathrm{km}/\mathrm{h}\phantom{\rule{0ex}{0ex}}{A}_{fast=}160\mathrm{km}/\mathrm{h}$

## Step 2: Determining the concept

This problem is based on the drag force which is a type of friction. This is the force acting opposite to the relative motion of an object moving with respect to the surrounding medium. Using the concept of the drag force and terminal speed the ratio of the effective cross-sectional area A in the slower position to that in the faster position.

Formula:

The terminal speed is given by

${v}_{t}=\sqrt{\frac{2{F}_{g}}{CpA}}$

Where C is the drag coefficient,${}^{p}$ is the fluid density, A is the effective cross-sectional area, and ${}^{{F}_{g}}$ is the gravitational force.

solve for the area

$A=\frac{2mg}{Cp{{v}^{2}}_{t}}$

which illustrates the inverse proportionality between the area and the speed-squared

## Step 3: Determining the ratio of the effective cross-sectional area A in the slower position to that in the faster position

When a ratio of areas has been setup of the slower case to the faster case,

$\frac{{A}_{slow}}{{A}_{fast}}={\left(\frac{110\mathrm{km}/\mathrm{h}}{160\mathrm{km}/\mathrm{h}}\right)}^{2}\phantom{\rule{0ex}{0ex}}=3.75$

Hence, the ratio of the effective cross-sectional area A in the slower position to that in the faster position is ${}^{3.75}$