Chapter 10.7, Problem 54AYU

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.