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 6 iron. The free-fall acceleration on the moon is 1/6 of its value on earth. Suppose he hits the ball with a speed of 27 m/s at an angle 10degree above the horizontal. How much farther did the ball travel on the moon than it would have on earth? Express your answer to two significant figures and include the appropriate units. For how much more time was the ball in flight? Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

Alright, the components of the ball's initial velocity are, for the horizontal, 27m/scos10°, and 27m/ssin10° for the vertical. 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(27m/ssin10°) / (9.8m/s² / 6)
= 5.74s (rounded)

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

x = (v + v)t / 2
= (27m/scos10° + 27m/scos10°)5.74s / 2
= 152.62m (rounded)

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

t = 2v / g
= 2(27m/ssin10°) / 9.8m/s²)
= 0.956s

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

x = (v + v)t / 2
= (27m/scos10° + 27m/scos10°)0.956s / 2
= 25.41m (rounded)

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.