Chapter 7.3, Problem 109AYU

Projectile Motion The horizontal distance that a projectile will travel in the air (ignoring air resistance) is given by the equation

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

where $v_0$ is the initial velocity of the projectile, $\theta$ is the angle of elevation, and $g$ is acceleration due to gravity ( $9.8$ meters per second squared).

a. If you can throw a baseball with an initial speed of $34.8$ meters per second, at what angle of elevation $\theta$ should you direct the throw so that the ball travels a distance of 107 meters before striking the ground?

b. Determine the maximum distance that you can throw the ball.

c. Graph $R=R\left(\theta \right)$, with $v_0=34.8$ meters per second.

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

To determine

To find:

a. If you can throw a baseball with an initial speed of $34.8$ meters per second, at what angle of elevation $\theta$ should you direct the throw so that the ball travels a distance of 107 meters before striking the ground?

Answer to Problem 109AYU

a. $\theta \approx 30°$

Explanation of Solution

Given:

The horizontal distance that a projectile will travel in the air (ignoring air resistance) is given by the equation

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

where $v_0$ is the initial velocity of the projectile, $\theta$ is the angle of elevation, and $g$ is acceleration due to gravity ($9.8$ meters per second squared).

Calculation:

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

$107=\frac{{34.8}^2\sin 2\theta }{9.8}$

$\sin 2\theta =0.87$

$2\theta ={\sin }^{-1}\left(0.87\right)$

$2\theta =60.46°$

$\theta \approx 30°$

To determine

To find:

b. Determine the maximum distance that you can throw the ball.

Answer to Problem 109AYU

b. $R\left(\theta \right)=123.6$ m.

Explanation of Solution

Given:

The horizontal distance that a projectile will travel in the air (ignoring air resistance) is given by the equation

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

where $v_0$ is the initial velocity of the projectile, $\theta$ is the angle of elevation, and $g$ is acceleration due to gravity ($9.8$ meters per second squared).

Calculation:

b. Maximum range $\sin 2\theta =1$.

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

$R\left(\theta \right)=\frac{{34.8}^2}{9.8}$

$R\left(\theta \right)=\mathrm{123.6m}$

To determine

To find:

c. Graph $R\text{ = }R\left(\theta \right)$, with $v_0=34.8$ meters per second.

Answer to Problem 109AYU

c.

Explanation of Solution

Given:

The horizontal distance that a projectile will travel in the air (ignoring air resistance) is given by the equation

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

where $v_0$ is the initial velocity of the projectile, $\theta$ is the angle of elevation, and $g$ is acceleration due to gravity ($9.8$ meters per second squared).

Calculation:

c.

To determine

To find:

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

Answer to Problem 109AYU

The results of (a) and (b) verified in (c).

Explanation of Solution

Given:

The horizontal distance that a projectile will travel in the air (ignoring air resistance) is given by the equation

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

where $v_0$ is the initial velocity of the projectile, $\theta$ is the angle of elevation, and $g$ is acceleration due to gravity ($9.8$ meters per second squared).

Calculation:

d. The results of (a) and (b) verified in (c).