Chapter 7.3, Problem 110AYU

Projectile Motion Refer to Problem 111.

a. 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?

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

c. Graph $R=R\left(\theta \right)$, with $v_0=40$ 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 40 meters per second, at what angle of elevation u should you direct the throw so that the ball travels a distance of 110 meters before striking the ground?

Answer to Problem 110AYU

a. $\theta \approx 21°=0.377$

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

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

$\sin 2\theta =0.67375$

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

$2\theta =42.35°$

$\theta \approx 21°=0.377$

To determine

To find:

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

Answer to Problem 110AYU

b. $R\left(\theta \right)=163.3$ 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 $\sin 2\theta =1$

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

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

$R\left(\theta \right)=\mathrm{163.3m}$

To determine

To find:

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

Answer to Problem 110AYU

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 110AYU

d. 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).