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fost5571 · Sep 24, 2009, 08:11 PM

Apollo astronauts took a "nine iron" to the Moon and hit a golf ball about 180 m! Assuming that the swing, launch angle, and so on, were the same as on Earth where the same astronaut could hit only 35m, estimate the acceleration due to gravity on the surface of the moon. (neglect air resistance)

Unknown008 · Sep 25, 2009, 12:07 PM

I'll give you the whole way to do it, taking the angle being the same in each case.

1. You know that the time of flight is given by the formula $v = u+at$

Well, since the initial velocity u is equal to -u, the final velocity of the ball, you have:

$-u = u +at$

$-2u = at$

$t = \frac{-2u}{a}$

The acceleration has a negative value, since I'm taking all upward direction as positive.

Then, the distance that the ball goes is given by $s = ut$

Replace t.

$s = u(\frac{-2u}{a})$

$s = \frac{-2u^2}{a}$

So, the distance 's' is inversely proportional to the acceleration which is due to gravity.

You can put it like this:

$s = \frac{k}{a}$

You have the value of s for Earth and you know the acceleration due to gravity on the Earth. Find 'k'.

Then, use the distance on the moon with 'k' to find 'a', the acceleration due to the gravity of the moon.

I hope it helped! :)