Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

A golfer tees off from a location precisely at = 28.0 degree north latitude. He

ID: 2129811 • Letter: A

Question

A golfer tees off from a location precisely at = 28.0 degree north latitude. He hits the ball due south, with range 290 m. The ball's initial velocity is at 48.0 degree above the horizontal. Suppose air resistance is negligible for the golf ball. 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 REcos i = (6.37 106 m) cos 28.0 degree, as shown in the figure. The tee completes one revolution each day Find the eastward speed of the tee, relative to the stars. The hole is also moving east, but it is 290 m farther south and thus at a slightly lower latitude f. Because the hole moves in a slightly larger circle, its speed must be greater than that of the tee. By how much does the hole's speed exceed that of the tee? During the time the ball is in flight, it moves both upward and downward as well as southward, 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). How far to the west of the hole does the ball land?

Explanation / Answer

a The ball will travel a distance equal to the flight time times the horizontal component of its velocity:
v cos(elevation) * t = range

Solve for your flight time:
t = range / (v cos(elevation) ) = 290/v cos 48

c) speed of hole = omega R, where R is the distance from the hole to the axis of the earth's rotation

The tee is going to be closer to the axis of rotation by a distance equal to the range times the sine of the lattitude.

speed of tee = omega (R - range sin(lattitude) )

So the difference in speed is omega range sin(lattitude)

Multiply that speed difference by the flight time to calculate the offset in landing position.

offset = deltav * t
= omega range^2 sin(lattitude) / (v cos(elevation) )

They give you the range, elevation, lattitude, and takeoff speed. Omega is 2pi radians per day. Plugnchug.

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote