Chapter 10.7, Problem 53AYU

To determine

To find:

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

Answer to Problem 53AYU

a. $x = \left(145\text{ cos}20\right)t$, $y = -16t² + 145t\text{sin}20 + 5$

Explanation of Solution

Given:

Ichiro throws a baseball with an initial speed of 145 feet per second at an angle of $20°$ to the horizontal. The ball leaves Ichiro’s hand at a height of 5 feet.

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:

$v_o = 145$, $\theta = 20°$

a. $x = \left(145\text{cos}20\right)t$

$y = -\frac{1}{2} \&*\#x00A0;32t² + (145\sin 20)t + 5$

$y = -16t² + 145t\text{sin}20 + 5$

To determine

To find:

b. How long is the ball in the air?

Answer to Problem 53AYU

b. $3.20$ secconds.

Explanation of Solution

Given:

Ichiro throws a baseball with an initial speed of 145 feet per second at an angle of $20°$ to the horizontal. The ball leaves Ichiro’s hand at a height of 5 feet.

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:

$v_o = 145$, $\theta = 20°$

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

$y = -16t² + 145t\text{sin}20 + 5$

$y = \left(-145\text{sin}20° \pm \sqrt{{\left(145\text{sin}20\right)}^2} - 4\left(-16\right)\left(5\right)\right)/2\left(-16\right)$

$y \approx -.10 \text{or} 3.2$

The ball is in the air for $3.20$ seconds.

To determine

To find:

c. The horizontal distance that the ball travels.

Answer to Problem 53AYU

c. 436 feet

Explanation of Solution

Given:

Ichiro throws a baseball with an initial speed of 145 feet per second at an angle of $20°$ to the horizontal. The ball leaves Ichiro’s hand at a height of 5 feet.

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:

$v_o = 145$, $\theta = 20°$

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

$x = 145 \&*\#x00A0;\text{cos}20 \&*\#x00A0;3.20$ $\approx \mathrm{ 436}$ feet.

To determine

To find:

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

Answer to Problem 53AYU

d. $y \approx 43.43$ feet.

Explanation of Solution

Given:

Ichiro throws a baseball with an initial speed of 145 feet per second at an angle of $20°$ to the horizontal. The ball leaves Ichiro’s hand at a height of 5 feet.

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:

$v_o = 145$, $\theta = 20°$

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

$t = -\frac{b}{2a} = \frac{145\sin \left(20°\right)}{2\left(-16\right)} \approx 1.55$ seconds.

Substituting the value in $y = -16t² + 145t\text{sin}20 + 5$

$y = -16{\left(1.55\right)}^2 + 145\left(1.55\right)\text{sin}20 + 5$

$y \approx 43.43$ feet.

To determine

To find:

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

Answer to Problem 53AYU

Explanation of Solution

Given:

Ichiro throws a baseball with an initial speed of 145 feet per second at an angle of $20°$ to the horizontal. The ball leaves Ichiro’s hand at a height of 5 feet.

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:

$v_o = 145$, $\theta = 20°$

e. Graph of the equation found in a.