Chapter 1, Problem 138RE

To determine

To calculate:The equation that depicts that the maximum range of the projectile $R$ varies directly to square of its velocity $v$ and use it to find out the maximum range if the ball is thrown at $70$ miles per hour and if it is given thatthemaximum range of $242$ feet is attained when ball is thrown with a velocity of $60$ miles per hour.

Answer to Problem 138RE

The maximum range if the ball is thrown at $70$ miles per hour is $R\approx 329.39$ feet

Explanation of Solution

Given information:

Here, the maximum range of the projectile $R$ varies directly to square of its velocity $v$

Also, the maximum range of $242$ feet is attained when ball is thrown with a velocity of $60$ miles per hour.

New velocity is $70$ miles per hour

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 maximum range of the projectile $R$ varies directly to square of its velocity $v$

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:

  $R\alpha v^2$

  $R=k{v}^2$ $\left(1\right)$

Where $k$ denotes the proportionality constant.

It is also given $R=242$ feet when $v=60$ miles per hour.

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

  $\begin{array}{l}242=k{\left(60\right)}^2\\ k=\frac{242}{3600}\\ k=\frac{121}{1800}\end{array}$

Therefore, the proportionality constant $k=\frac{121}{1800}$

Therefore, $\left(1\right)$ reduces to $R=\frac{121}{1800}{v}^2$

Now for new velocity of $v=70$ miles per hour.

Replace the value of $k=\frac{121}{1800}$ and $v=70$ in $\left(1\right)$ to get:

  $\begin{array}{l}R=\frac{121}{1800}{\left(70\right)}^2\\ R=\frac{121}{1800}\left(4900\right)\\ R\approx 329.39\end{array}$

Thus, the maximum range if the ball is thrown at $70$ miles per hour is $R\approx 329.39$ feet