Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
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Chapter 4, Problem 33PQ

Dock diving is a great form of athletic competition for dogs of all shapes and sizes (Fig. P4.33). Sheba, the American Pit Bull Terrier, runs and jumps off the dock with an initial speed of 9.02 m/s at an angle of 25° with respect to the surface of the water. a. If Sheba begins at a height of 0.84 m above the surface of the water, through what horizontal distance does she travel before hitting the surface of the water? b. Write an expression for the velocity of Sheba, in component form, the instant before she hits the water. c. Determine the peak height above the water reached by Sheba during her jump.

Chapter 4, Problem 33PQ, Dock diving is a great form of athletic competition for dogs of all shapes and sizes (Fig. P4.33).

FIGURE P4.33

(a)

Expert Solution
Check Mark
To determine

The horizontal distance travelled by Sheba before hitting the surface of the water.

Answer to Problem 33PQ

The horizontal distance travelled by Sheba before hitting the surface of the water is 7.8 m.

Explanation of Solution

Sheba, the dog, is dock diving in an athletic competition for dogs and jumps off the dock with an initial speed of vi=9.02 m/s at an angle 25° with respect to the surface of the water. The initial height of Sheba is yi=0.84 m above the surface of the water.

Write the formula for the vertical position

    yf=yi+vyit+12ayt2                                                                                            (I)

Here, yf is the final position in y axis, yi is the initial position in y axis, ay is the vertical acceleration, t is the time and vyi is the initial vertical velocity.

Write the formula for the horizontal position

    xf=xi+vxit                                                                                                       (II)

Here, xf is the final position in x axis, xi is the initial position in x axis and vxi is the horizontal velocity

Sheba jumped from initial position yi=0.84 m. The initial vertical velocity of Sheba is

vyf=visinθ=(9.02 m/s)sin(25°)=3.81 m/s.

The vertical acceleration is due to gravity pulling the Sheba in direction downward, ay=9.81 m/s2.

Substituting 0.84 m for yi, 3.81 m/s for vyi and 9.81 m/s2 for ay in equation (I) to find the value of vyi

0=0.84 m+(3.81 m/s)t(9.81 m/s2)t

Solving the quadratic equation, the valuable root is t=0.96 s.

The time taken for Sheba to reach the water surface is 0.96 s. The horizontal velocity of Sheba is

 vxi=vicosθ=(9.02 m/s)cos(25°)=8.17 m/s

Substitute 8.17 m/s for vxi and 0.96 s for t in equation (III) to find the value of Δx

Δx=vxitΔx=(8.17 m/s)(0.96 s)Δx=7.8 m

Thus, the horizontal distance travelled by Sheba before hitting the surface of the water is 7.8 m.

(b)

Expert Solution
Check Mark
To determine

Write the velocity of Sheba at the instant before she hits the water in component form.

Answer to Problem 33PQ

The velocity of Sheba at the instant before she hits the water in component form is v=(8.2i^5.6j^) m/s.

Explanation of Solution

Write the formula for the velocity vector equation for the projectile motion

    v=(vicosθ)i^+(visinθgt)j^                                                                              (III)

Here, vi is the initial velocity, θ is the angle, g is the acceleration due to gravity and v is the velocity vector.

The horizontal velocity component is vx=vicosθ=(9.02 m/s2)cos(25°)=8.2 m/s.

The vertical velocity component is vy=visinθgt=(9.02 m/s2)sin(25°)(9.8 m/s2)(0.96 s)=5.6 m/s.

Substituting in equation (III) to find the value of v

v=(8.2i^5.6j^) m/s

Thus, the velocity of Sheba at the instant before she hits the water in component form is v=(8.2i^5.6j^) m/s.

(c)

Expert Solution
Check Mark
To determine

The peak height above the water surface reached by Sheba during her jump.

Answer to Problem 33PQ

The peak height above the water surface reached by Sheba during her jump is 1.58 m.

Explanation of Solution

Write the formula for the vertical velocity component for the projectile motion

    vy=vyfgt                                                                              (IV)

The initial vertical velocity of Sheba is vyf=visinθ=(9.02 m/s)sin(25°)=3.81 m/s. The vertical velocity of Sheba at maximum height is zero, vy=0.

Substituting 3.81 m/s for vyf and 9.81 m/s2 for g in equation (I) to find the value of t

0=3.81 m/s(9.8 m/s2)tt=3.819.8 st=0.39 s

The time taken for Sheba to reach the peak height is 0.39 s.

Substituting 0.84 m for yi, 3.81 m/s for vyi, 0.39 s for t  and 9.81 m/s2 for ay in equation (I) to find the value of yf

yf=0.84 m+(3.81 m/s)(0.39 s)12(9.81 m/s2)(0.39 s)2yf=1.58 m

Thus, the peak height above the water surface reached by Sheba during her jump is 1.58 m.

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Chapter 4 Solutions

Physics for Scientists and Engineers: Foundations and Connections

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