Chapter 7.3, Problem 110AYU

a.

To determine

What angle of elevation $\theta$ should you direct the throw a baseball.

Answer to Problem 110AYU

  $\boxed{\theta =21.18°,42.36°}$

Explanation of Solution

Given information:

If you can throw a baseball with an initial speed of $40$ meters per second, at what angle of elevation $\theta$ should you direct the throw so that the ball travels a distance of $110$ meters before striking the ground?

Calculation:

The range of the projectile is given as

  $R\left(\theta \right)=\frac{v_0{}^2\sin (2\theta )}{g}\mathrm{......}(1)$

We have to find out the angle of elevation such that the ball travels a distance of $110$ meters before striking the ground.

Consider the values

  $v_0=40m/s$ and $g=9.8m/s_2$ .

Let us put these values in equation $R\left(\theta \right)=\frac{v_0{}^2\sin (2\theta )}{g}\mathrm{......}(1)$ .

  $R\left(\theta \right)=\frac{v_0{}^2\sin (2\theta )}{g}$

  $110=\frac{{(40)}^2\sin (2\theta )}{9.8}$

  $\sin (2\theta )=\frac{\left(110\right)\left(9.8\right)}{{\left(40\right)}^2}$

  $\sin \left(2\theta \right)=0.673$

Hence, solving this equation, we get

  $2\theta =42.36°,137.64°$

Or

  $\boxed{\theta =21.18°,42.36°}$

To determine

Determine the maximum distance that you can throw the ball.

Answer to Problem 110AYU

  $\boxed{R_{\max }(\theta )=163.3meters}$

Explanation of Solution

Given information:

Determine the maximum distance that you can throw the ball.

Calculation:

The range of the projectile is given as

  $R\left(\theta \right)=\frac{v_0{}^2\sin (2\theta )}{g}\mathrm{......}(1)$

Throwing maximum distance means maximum range ,that is maximum value of $R(\theta )$ in equation $(1)$ . As $v_0$ and $g$ are constant we can only maximize $\sin (2\theta )$ . And we know that maximum value of sine function is 1. Hence, we have

  $R_{\max }(\theta )=\frac{v_0{}^2\sin (2\theta )}{g}$

  $=\frac{{(40)}^2(1)}{9.8}$

  $\boxed{R_{\max }(\theta )=163.3meters}$

To determine

Plot the graph.

Answer to Problem 110AYU

  

Explanation of Solution

Given information:

Graph $R=R(\theta )$ , with $v_0=40$ meters per second.

Calculation:

The range of the projectile is given as

  $R\left(\theta \right)=\frac{v_0{}^2\sin (2\theta )}{g}\mathrm{......}(1)$

Figure below shows the graph of $R=R(\theta )$ with $v_0=40$ m/s.

  $R\left(\theta \right)=\frac{v_0{}^2\sin (2\theta )}{g}$

To determine

Verify the results obtained in parts (a) and (b).

Answer to Problem 110AYU

The results in part (a) and (b) are verified.

Explanation of Solution

Given information:

Verify the results obtained in parts (a) and (b) using a graphing utility.

Calculation:

The range of the projectile is given as

  $R\left(\theta \right)=\frac{v_0{}^2\sin (2\theta )}{g}\mathrm{......}(1)$

We can observe from the graph that range of $110$ meters can be achieved at an angle of $21.18°$ and $42.36°$ , and maximum range is approximately $163.3$ meters.

Here we use $TI-83$ calculator to graph the function of range and find the maximum distance.

First insert the equation $R\left(\theta \right)=\frac{v_0{}^2\sin (2\theta )}{g}$ with $v_0=40m/s$ and $g=9.8m/s^2$ that is $R\left(\theta \right)=\frac{v_0{}^2\sin (2\theta )}{g}$

  $=\frac{{(40)}^2\sin (2\theta )}{9.8}$

  $=163.3\sin (2\theta )$

  

Now press WINDOW key and set the range of viewing window as.

  

By pressing the GRAPH key we get

  

Now for finding the maximum point we use the following process:

Press 2^nd key and then TRACE key.

Press 4^th key to select MAXIMUM.

Move the cursor to the left of the maximum point, then press ENTER key to select left bound, now move the cursor to the right of the maximum point, then press ENTER key to select right bound now by pressing the ENTER key third time we get the maximum point as

  $x=0.785,y=163.3$

  

Here we see maximum range is

  $y=163.3$

Also by using the VALUE feature we get that range of $110$ meters can be achieved at an angle of $21.18°$ and $42.36°$ .

Hence, the results in part (a) and (b) are verified.