Answer a. x = ( 145 cos 20 ) t , y = − 16 t ² + 145 t sin 20 + 5 Explanation 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 = ( v o cos θ ) t , y = − 1 2 g t ² + ( v o sin θ ) t + h Calculation: v o = 145 , θ = 20 ° a. x = ( 145 cos 20 ) t y = − 1 2 &*#x00A0; 32 t ² + ( 145 sin 20 ) t + 5 y = − 16 t ² + 145 t sin 20 + 5 To determine To find: b. How long is the ball in the air? Answer b. 3.20 secconds. Explanation 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 = ( v o cos θ ) t , y = − 1 2 g t ² + ( v o sin θ ) t + h Calculation: v o = 145 , θ = 20 ° b. The ball is in the air until y = 0 , Solve: y = − 16 t ² + 145 t sin 20 + 5 y = ( − 145 sin 20 ° ± ( 145 sin 20 ) 2 − 4 ( − 16 ) ( 5 ) ) / 2 ( − 16 ) y ≈ − .10 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 c. 436 feet Explanation 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 = ( v o cos θ ) t , y = − 1 2 g t ² + ( v o sin θ ) t + h Calculation: v o = 145 , θ = 20 ° c. Horizontal distance x = ( v o cos θ ) t x = 145 &*#x00A0; cos 20 &*#x00A0; 3.20 ≈ 436 feet. To determine To find: d. When is the ball at its maximum height? Determine the maximum height of the ball. Answer d. y ≈ 43.43 feet. Explanation 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 = ( v o cos θ ) t , y = − 1 2 g t ² + ( v o sin θ ) t + h Calculation: v o = 145 , θ = 20 ° d. The maximum height is at vertex of the quadratic function. t = − b 2 a = 145 sin ( 20 ° ) 2 ( − 16 ) ≈ 1.55 seconds. Substituting the value in y = − 16 t ² + 145 t sin 20 + 5 y = − 16 ( 1.55 ) 2 + 145 ( 1.55 ) sin 20 + 5 y ≈ 43.43 feet. To determine To find: e. Using a graphing utility, simultaneously graph the equations found in part (a). Answer Explanation 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 = ( v o cos θ ) t , y = − 1 2 g t ² + ( v o sin θ ) t + h Calculation: v o = 145 , θ = 20 ° e. Graph of the equation found in a.

9th Edition

ISBN: 9780321716835

Author: Michael Sullivan

Publisher: Addison Wesley

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Question

Chapter 10.7, Problem 53AYU

To determine

To find:

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

Expert Solution

Answer to Problem 53AYU

a. $x = (145 cos 20) t$ , $y = - 16 t ² + 145 t sin 20 + 5$

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:

$x = (v_{o} cos θ) t$ , $y = - \frac{1}{2} g t ² + (v_{o} sin θ) t + h$

Calculation:

$v_{o} = 145$ , $θ = 20 °$

a. $x = (145 cos 20) t$

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

$y = - 16 t ² + 145 t sin 20 + 5$

To determine

To find:

b. How long is the ball in the air?

Expert Solution

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:

$x = (v_{o} cos θ) t$ , $y = - \frac{1}{2} g t ² + (v_{o} sin θ) t + h$

Calculation:

$v_{o} = 145$ , $θ = 20 °$

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

$y = - 16 t ² + 145 t sin 20 + 5$

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

$y \approx - .10 or 3.2$

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

To determine

To find:

c. The horizontal distance that the ball travels.

Expert Solution

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:

$x = (v_{o} cos θ) t$ , $y = - \frac{1}{2} g t ² + (v_{o} sin θ) t + h$

Calculation:

$v_{o} = 145$ , $θ = 20 °$

c. Horizontal distance $x = (v_{o} cos θ) t$

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

To determine

To find:

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

Expert Solution

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:

$x = (v_{o} cos θ) t$ , $y = - \frac{1}{2} g t ² + (v_{o} sin θ) t + h$

Calculation:

$v_{o} = 145$ , $θ = 20 °$

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

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

Substituting the value in $y = - 16 t ² + 145 t sin 20 + 5$

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

$y \approx 43.43$ feet.

To determine

To find:

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

Expert Solution

Answer to Problem 53AYU

Precalculus, Chapter 10.7, Problem 53AYU , additional homework tip 1

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:

$x = (v_{o} cos θ) t$ , $y = - \frac{1}{2} g t ² + (v_{o} sin θ) t + h$

Calculation:

$v_{o} = 145$ , $θ = 20 °$

e. Graph of the equation found in a.

Precalculus, Chapter 10.7, Problem 53AYU , additional homework tip 2

Chapter 10.7, Problem 52AYU

Chapter 10.7, Problem 54AYU

Chapter 10 Solutions

Precalculus

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

Solution Summary: The author explains how Ichiro throws a baseball with an initial speed of 145 feet per second at an angle of 20 ° to the horizontal.

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

Answer to Problem 53AYU

Explanation of Solution

To find: b. How long is the ball in the air?

Answer to Problem 53AYU

Explanation of Solution

To find: c. The horizontal distance that the ball travels.

Answer to Problem 53AYU

Explanation of Solution

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

Answer to Problem 53AYU

Explanation of Solution

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

Answer to Problem 53AYU

Explanation of Solution

Chapter 10 Solutions

Additional Math Textbook Solutions

To find:

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

To find:

b. How long is the ball in the air?

To find:

c. The horizontal distance that the ball travels.

To find:

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

To find:

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