On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball wit

Question

On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a golf club improvised from a tool. The free-fall acceleration on the moon is 1/6of its value on earth. Suppose he hit the ball with a speed of 23 m/s at an angle 34 above the horizontal.

Part A

How long was the ball in flight?

Express your answer to two significant figures and include the appropriate units.

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Part B

How far did it travel?

Express your answer to two significant figures and include the appropriate units.

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Part C

Ignoring air resistance, how much farther would it travel on the moon than on earth?

Express your answer to two significant figures and include the appropriate units.

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Explanation / Answer

Part A -

Here Horiontal component of the velocity, Vx = 23 * cos34°,

And vertical component of the velocity Vy = 23 * sin34°

The final speed in the horizontal is the same as the initial because there is zero acceleration in that direction. Since we can assume that the ball lands vertically at the same height it was hit from, its displacement is zero as well.

The time of flight is:

y = vt + 0.5gt²

Since we determined that vertical displacement is zero, this becomes:

0 = vt + 0.5gt²
gt² = 2vt
t = 2v / g
= 2*(23 * sin34°) / (9.8m/s² / 6)
= 15.75 s = 16.0 s

Part B -

The distance it traveled may be found from the horizontal components of the ball's velocity:

x = (v + v)t / 2
= (23m/s*cos34° + 23m/s*cos34°)16s / 2
= ( 19.07 + 19.07)*8 = 305.12 m = 305.0 m

Part C -

On earth, the time of flight will be different, it is:

t = 2v / g
= 2(23m/s*sin34°) / 9.8m/s²)
= 2.6 s

So, by the same method as before, its range is:

x = (v + v)t / 2
= (23m/s*cos34° + 23m/s*cos34°)2.6s / 2
= (19.07 + 19.07)*1.3 = 50.0 m

So, Xmoon - Xearth = 305 - 50 = 255.0 m

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