Chapter 1, Problem 137RE

To determine

To calculate:The equation that depicts that the terminal velocity of parachutist $v$ varies directly with respect to the square root of his weight $w$ and use it to find out the terminal velocity when the parachutist weighs $240$ lbs. if the $160$ lbs. parachutist attains a terminal velocity of $9$ miles per hour.

Answer to Problem 137RE

The terminal velocity when the parachutist weighs $240$ lbs. is $v\approx 11.02$ miles per hour.

Explanation of Solution

Given information:

Here, the terminal velocity of parachutist $v$ varies directly with respect to the square root of his weight $w$

Also, $160$ lbs. parachutist attains a terminal velocity of $9$ miles per hour.

New weight is $240$ lbs.

Formula used:

For 2 variables say, $x$ and $y$ , the statement $x$ is directly proportional to $y$ can be written as:

  $x\alpha y$

Which can be written as:

  $x=ky$

Where $k$ denotes the proportionality constant.

Similarly the statement $x$ is inversely proportional to $y$ can be interpreted as:

  $x\alpha \frac{1}{y}$

Which can be written as:

  $x=k\left(\frac{1}{y}\right)$

Where $k$ denotes the proportionality constant.

Calculation:

As the terminal velocity of parachutist $v$ varies directly with respect to the square root of his weight $w$

For 2 variables say, $x$ and $y$ , the statement $x$ is directly proportional to $y$ can be written as:

  $x\alpha y$

Which can be written as:

  $x=ky$

Where $k$ denotes the proportionality constant.

Hence, this variation can be expressed as follows:

  $v\alpha \sqrt{w}$

  $v=k\sqrt{w}$ $\left(1\right)$

Where $k$ denotes the proportionality constant.

It is also given $v=9$ miles per hour when $w=160$ lbs.

Put these values in $\left(1\right)$ to get:

  $\begin{array}{l}9=k\sqrt{160}\\ 9=k\sqrt{4*4*10}\\ 9=k4\sqrt{10}\\ k=\frac{9}{4\sqrt{10}}\end{array}$

Therefore, the proportionality constant $k=\frac{9}{4\sqrt{10}}$

Therefore, $\left(1\right)$ reduces to $v=\frac{9}{4\sqrt{10}}\sqrt{w}$

Now for new weight of $w=240$ lbs.

Replace the value of $k=\frac{9}{4\sqrt{10}}$ and $w=240$ in $\left(1\right)$ to get:

  $\begin{array}{l}v=\frac{9}{4\sqrt{10}}\sqrt{240}\\ v=\frac{9}{4\sqrt{10}}4\sqrt{15}\\ v\approx 11.02\end{array}$

Thus, the terminal velocity when the parachutist weighs $240$ lbs. is $v\approx 11.02$ miles per hour.