Chapter 10.7, Problem 53AYU

Projectile Motion Ichiro throws a baseball with an initial speed of 145 feet per second at an angle of ${20}^{\circ }$ to the horizontal. The ball leaves Ichiro’s hand at a height of 5 feet.

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

b. How long is the ball in the air?

c. Determine the horizontal distance that the ball travels.

d. When is 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 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.