TPrj13

Alan Shepard was the first moon golfer. If he could hit an iron shot 140 yards on earth, how far would that shot travel on the moon where the "gravity" is 1/6th that of the earth? Assume that on earth he launches the ball at 45° with a certain velocity and that he achieves the same values of angle and velocity on the moon.

The first step in the problem is to determine the initial velocity on the earth.

Set up the equations.

We have the range on earth as 140 yds = 420 ft. So write

When y = 0 the quadratic in t should give two answers (for t); 0 and the time of flight.

The values of t that make y = 0 are

Substitute this latter (non-zero) term into the previous equation.

Use this v₀ for the “moon shot”. Set up the system with u and v to help keep the coordinates separate. Remember: “g” on the moon is 32/6 = 5.6 ft/s².

Solve v for t when v = 0

Substitute this t into the u equation to find the range.

Because there is much less “air” on the moon, the air resistance is nearly eliminated and his shot would be longer than the calculated 800 yd.