Apollo 14 Golf on Moon (Projectile Motion)

May 2008
1
0
During the Apollo 14 mission, American astronaut Alan Shepard hit a golf ball. Given that the acceleration due to gravity on the moon's surface is 1.62 m/s^2, assuming that the angle of the shot was 20 degrees above the surface, and that the ball traveled 350 m, calculate the initial velocity of the shot.

I got 29.7m/s, but my teacher said the correct answer should be 18.3m/s. Which one is correct? Help!
 
May 2008
7
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Are you sure there are no more values given? It would help to see your workings. Rob
 

topsquark

Forum Staff
Apr 2008
3,115
661
On the dance floor, baby!
During the Apollo 14 mission, American astronaut Alan Shepard hit a golf ball. Given that the acceleration due to gravity on the moon's surface is 1.62 m/s^2, assuming that the angle of the shot was 20 degrees above the surface, and that the ball traveled 350 m, calculate the initial velocity of the shot.

I got 29.7m/s, but my teacher said the correct answer should be 18.3m/s. Which one is correct? Help!
I agree with you.

Let the initial speed be \(\displaystyle v_0\). Call +x horizontal to the Moon's surface and in the direction of the shot, and call +y perpendicular to the Moon's surface and upward.

Then
\(\displaystyle x = x_0 + v_{0x}t\)

\(\displaystyle 350 = v_0t~cos(20)\)

and
\(\displaystyle y = y_0 + v_{0y}t + \frac{1}{2}a_yt^2\)

\(\displaystyle 0 = v_0t~sin(20) - 0.81t^2\)

Using the x equation to solve for t we get
\(\displaystyle t = \frac{350}{v_0~cos(20)}\)
and inserting this value of t into the y equation we get
\(\displaystyle 0 = v_0 \left ( \frac{350}{v_0~cos(20)} \right ) ~sin(20) - 0.81 \left ( \frac{350}{v_0~cos(20)} \right ) ^2\)

Solving for \(\displaystyle v_0\) I get 29.7 m/s as well.

-Dan