Chapter 10.7, Problem 54AYU

Projectile Motion Mark Texeira hit a baseball with an initial speed of 125 feet per second at an angle of ${40}^{\circ }$ to the horizontal. The ball was hit at a height of 3 feet off the ground.

a. Find parametric equations that model the position of the ball as a function of time.

b. How long was the ball in the air?

c. Determine the horizontal distance that the ball traveled.

d. When was the ball at its maximum height? Determine the maximum height of the ball.

e. Using a graphing utility, simultaneously graph the equations found in part (a).

To determine

To find:

a. Parametric equations that model the position of the ball as a function of time.

Answer to Problem 54AYU

a. $x = \left(125\text{cos}40\right)t$, $y = -16t² + 125t\text{sin}40 + 3$

Explanation of Solution

Given:

Mark Texeira hit a base ball with an initial speed of 125 feet per second at an angle of $40°$ to the horizontal. The ball was hit at a height of 3 feet off the ground.

Formula used:

$x = \left(v_o\text{cos}\theta \right)t$, $y = -\frac{1}{2}gt² + \left(v_o\text{sin}\theta \right)t + h$

Calculation:

$v_o = 125$, $\theta = 40°$

a. $x = \left(125\text{cos}40\right)t$

$y = -16t² + 125t\text{sin}40 + 3$

To determine

To find:

b. How long is the ball in the air?

Answer to Problem 54AYU

b. $5.059$ secconds.

Explanation of Solution

Given:

Mark Texeira hit a base ball with an initial speed of 125 feet per second at an angle of $40°$ to the horizontal. The ball was hit at a height of 3 feet off the ground.

Formula used:

$x = \left(v_o\text{cos}\theta \right)t$, $y = -\frac{1}{2}gt² + \left(v_o\text{sin}\theta \right)t + h$

Calculation:

$v_o = 125$, $\theta = 40°$

b. The ball is in the air until $y = 0$, Solve:

$y = -16t² + 125t\text{sin}40 + 3$

$t = \frac{-125\sin 40 \pm \sqrt{{\left(125\text{sin}40\right)}^2 - 4\left(-16\right)\left(3\right)}}{2\left(-16\right)}$

$t \approx -0.037$ or $5.059$

The ball is in the air for $5.059$ seconds.

To determine

To find:

c. The horizontal distance that the ball travels.

Answer to Problem 54AYU

c. $484.41$ feet.

Explanation of Solution

Given:

Mark Texeira hit a base ball with an initial speed of 125 feet per second at an angle of $40°$ to the horizontal. The ball was hit at a height of 3 feet off the ground.

Formula used:

$x = \left(v_o\text{cos}\theta \right)t$, $y = -\frac{1}{2}gt² + \left(v_o\text{sin}\theta \right)t + h$

Calculation:

$v_o = 125$, $\theta = 40°$

c. Horizontal distance $x = \left(v_o\text{cos}\theta \right)t$

$\approx 484.41$ feet.

To determine

To find:

d. When is the ball at its maximum height? Determine the maximum height of the ball.

Answer to Problem 54AYU

d. $y \approx 103.87$ feet.

Explanation of Solution

Given:

Mark Texeira hit a base ball with an initial speed of 125 feet per second at an angle of $40°$ to the horizontal. The ball was hit at a height of 3 feet off the ground.

Formula used:

$x = \left(v_o\text{cos}\theta \right)t$, $y = -\frac{1}{2}gt² + \left(v_o\text{sin}\theta \right)t + h$

Calculation:

$v_o = 125$, $\theta = 40°$

d. The maximum height is at vertex of the quadratic function.

$t = -\frac{b}{2a} = - \frac{125\sin \left(40°\right)}{2\left(-16\right)} \approx 2.51$ seconds.

Substituting the value in $y = -16t² + 145t\text{sin}20 + 3$

$y = -16{\left(2.51\right)}^2 + 125\left(2.51\right)\text{sin}40 + 3$

$y \approx 103.87$ feet.

To determine

To find:

e. Using a graphing utility, simultaneously graph the equations found in part (a).

Answer to Problem 54AYU

Explanation of Solution

Given:

Mark Texeira hit a base ball with an initial speed of 125 feet per second at an angle of $40°$ to the horizontal. The ball was hit at a height of 3 feet off the ground.

Formula used:

$x = \left(v_o\text{cos}\theta \right)t$, $y = -\frac{1}{2}gt² + \left(v_o\text{sin}\theta \right)t + h$

Calculation:

$v_o = 125$, $\theta = 40°$

e. Graph of the equation found in a.