Consider a golfer here in Oregon, at exactly 45.0O latitude, hitting a golf ball

Question

Consider a golfer here in Oregon, at exactly 45.0O latitude, hitting a golf ball toward the hole, 275 m due south. The balls initial velocity is 48.0O above the horizontal. The golf ball is initially headed straight toward the hole for a holein- one, and would be a hole-in-one, except, the Earth is rotating!

a) Assume negligible air resistance and the tee and hole are at the same level. Solve the projectile problem and find the time the ball is in the air. (A review problem; nothing rotational here. Yet.)

b) Using radius of the Earth = 6.37x106 m, and a rotational period of 24 hrs, find the eastwardly-directed tangential speed of the tee. (Note that the ball will also have this initial eastwardly-directed speed.) Note that the radius of the tees circular path on the rotating Earth is not the radius of the Earth.

c) Since the hole is further south, it will have a greater eastwardly-directed tangential speed. Calculate how much greater the (tangential) speed of the hole is relative to the tee (and the flying ball).

d) Since the hole is moving to the east at a higher speed than the ball is moving to the east, the ball might miss the hole. By how much will that ball be deflected to the west when it reaches the hole?

Explanation / Answer

Hence "hooks" and "slices" which are the effects of the spin of the ball making small variations in wind resistance and large variations in the horizontal trajectory.

In the same way the ball is hit at perhaps 10 degrees from the horizontal but the lift from the spin keeps it in the air until the ball slows down. So it stays in the air a LOT longer than these figures suggest.

Of course if you did the golf on the moon then we could ignore wind resistance. But the moon is not spinning as fast as the earth and the gravity is less so the effects would be different.

If you ignore reality and physics then you could calculate the following.
Vertical component = V sin(48)
Horizontal component = V cos(48)
t = 2 V sin(48) / g
and Vcos(48) * t = 275
hence 2 V^2 cos(48) * sin*48) / g = 275
from this you can find V and t

If you find the rotational speed of the earth at 45 degrees of latitude and use interpolation to find the speed 275 m south of this you have the DIFFERENCE in the two speeds.
(V = 2 pi()R * cos(45) / (24*60*60) )
dV/ dx = dV/d theta * d theta /dx

If you assume that the speed difference increases linearly then you can answer all the remaining questions easily.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.