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