Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)
Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)
9th Edition
ISBN: 9781305266292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 6, Problem 67CP

A golfer tees off from a location precisely at ϕi = 35.0° north latitude. He hits the ball due south, with range 285 m. The ball’s initial velocity is at 48.0° above the horizontal. Suppose air resistance is negligible for the golf ball. (a) For how long is the ball in flight? The cup is due south of the golfer’s location, and the golfer would have a hole-in-one if the Earth were not rotating. The Earth’s rotation makes the tee move in a circle of radius RE cos ϕi = (6.37 × 106 m) cos 35.0° as shown in Figure P6.47. The tee completes one revolution each day. (b) Find the eastward speed of the tee relative to the stars. The hole is also moving cast, but it is 285 m farther south and thus at a slightly lower latitude ϕf. Because the hole moves in a slightly larger circle, its speed must he greater than that of the tee. (c) By how much does the hole’s speed exceed that of the tee? During the time interval the ball is in flight, it moves upward and downward as well as southward with the projectile motion you studied in Chapter 4, but it also moves eastward with the speed you found in part (b). The hole moves to the east at a faster speed, however, pulling ahead of the ball with the relative speed you found in part (c). (d) How far to the west of the hole does the ball land?

Figure P6.47

Chapter 6, Problem 67CP, A golfer tees off from a location precisely at i = 35.0 north latitude. He hits the ball due south,

(a)

Expert Solution
Check Mark
To determine

The time for which the ball be in flight.

Answer to Problem 67CP

The time for which the ball be in flight is 8.04sec .

Explanation of Solution

Given information: The range of the motion after hitting the ball is 285m . The direction of initial velocity of ball is 48.0° above the horizontal, the radius of the circular path is REcosϕi=(6.37×106m)cos35.0° and the location from which the golfer tees off is 35.0° north latitude.

The range of the parabolic motion is given as,

R=V0cosβt (I)

  • V0 is the initial velocity.
  • β is the direction of initial velocity.
  • t is the time.

The equation for parabolic motion is given as,'

y=y0+V0sinβtgt22yy0=V0sinβtgt22

As initial and final distance is equal, yy0=0 .

V0sinβtgt22=0V0sinβt=gt22

Rearrange the above expression for t .

t=2V0sinβg (II)

Substitute 2V0sinβg for t in equation (I).

R=V0cosβ(2V0sinβg)R=2V02cosβsinβgR=V02sin2βg

Rearrange the above expression for V0 .

V0=gRsin2β

Substitute gRsin2β for V0 in equation (II).

t=2(gRsin2β)sinβg

Substitute 285m for R , 9.8m/s2 for g and 48.0° for β in above expression.

t=2((9.8m/s2)(285m)sin2(48.0°))sin(48.0°)9.8m/s2=8.04sec

Conclusion: Therefore, the time for which the ball be in flight is 8.04sec .

(b)

Expert Solution
Check Mark
To determine

The eastward speed of the tee relative to the stars.

Answer to Problem 67CP

The eastward speed of the tee relative to the stars is 379m/s .

Explanation of Solution

Given information: The range of the motion after hitting the ball is 285m . The direction of initial velocity of ball is 48.0° above the horizontal, the radius of the circular path is REcosϕi=(6.37×106m)cos35.0° and the location from which the golfer tees off is 35.0° north latitude.

Formula to calculate the eastward speed of the tee relative to the stars is,

V=2πrT

Substitute (6.37×106m)cos35.0° for r and 24h for T in above expression.

V=2(3.14)((6.37×106m)cos35.0°)24h(3600sec1h)=379.2710m/s379m/s

Conclusion:

Therefore, the eastward speed of the tee relative to the stars is 379m/s .

(c)

Expert Solution
Check Mark
To determine

The value by which the hole's speed exceed that of the tee.

Answer to Problem 67CP

The value by which the hole's speed exceed that of the tee is 1.19×102m/s .

Explanation of Solution

Given information: The range of the motion after hitting the ball is 285m . The direction of initial velocity of ball is 48.0° above the horizontal, the radius of the circular path is REcosϕi=(6.37×106m)cos35.0° and the location from which the golfer tees off is 35.0° north latitude.

Formula to calculate the length of the arc is,

R=REϕ

Rearrange the above expression for ϕ .

ϕ=RRE(360°2π)

Substitute 285m for R and 6.37×106m for RE in above equation.

ϕ=285m6.37×106m(360°2(3.14))=0.0025°

Formula to calculate the speed of the hole is,

Vhole=2πREcos(35.0°ϕ)24h

Substitute 6.37×106m for RE , 0.0025° for ϕ in above expression.

Vhole=2(3.14)(6.37×106m)cos(35.0°0.0025°)24h((3600sec1h))=379.2829m/s

Calculate the difference between the speed of tee and the speed of hole.

ΔV=VholeV

Substitute 379.276m/s for Vhole and 379.271m/s for V in above expression.

ΔV=379.2829m/s379.2710m/s=1.19×102m/s

Conclusion:

Therefore, the value by which the hole's speed exceed that of the tee is 1.19×102m/s .

(d)

Expert Solution
Check Mark
To determine

The distance to the west of the hole from the position where the ball land.

Answer to Problem 67CP

The distance to the west of the hole from the position where the ball land is 9.55cm .

Explanation of Solution

Given information: The range of the motion after hitting the ball is 285m . The direction of initial velocity of ball is 48.0° above the horizontal, the radius of the circular path is REcosϕi=(6.37×106m)cos35.0° and the location from which the golfer tees off is 35.0° north latitude.

The distance to west of the hole is given as,

d=ΔV(t)

Substitute 1.19×102m/s for ΔV and 8.04sec for t in above expression.

d=1.19×102m/s(8.04sec)=9.55cm

Conclusion:

Therefore, the distance to the west of the hole from the position where the ball land is 9.55cm .

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Chapter 6 Solutions

Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)

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