Chapter 5.3, Problem 92E

To determine

To find: $\theta$ when the range $r$ of a projectile is given by $r=\frac{1}{32}{v}_0{}^2\sin 2\theta$ .

Answer to Problem 92E

  $\theta =53.1°$ .

Explanation of Solution

Given information:

A baseball is hit at an angle of $\theta$ with the horizontal and with an initial velocity of $v_0=100$ feet per second. An outfielder catches the ball 300 feet from home plate.

  

Calculation:

Let us consider the following figure

  

Let us consider that,

  $r=\frac{1}{32}{v}_0{}^2\sin 2\theta$

Here $r$ is the range in feet and $\theta$ is the angle with horizontal.

Let us find the value of $\theta$ for $r=300$ and $v_0=100$ ,

  $\begin{array}{c}r=\frac{1}{32}{v}_0{}^2\sin 2\theta \\ 300=\frac{1}{32}{\left(100\right)}^2\sin 2\theta \\ \sin 2\theta =0.96\\ 2\theta =\arcsin \left(0.96\right)\approx 73.74°\\ \theta \approx 36.9°\end{array}$

Or

  $\begin{array}{c}2\theta =180°-\arcsin \left(0.96\right)\approx 106.26°\\ \theta =53.1°\end{array}$