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 1/6 of its value on earth. Suppose he hit the ball with a speed of 35 m/s at an angle 27 ∘ above the horizontal. Part A How long was the ball in flight? Express your answer using two significant figures.

Question

Newt

Physics

30 October, 20:43

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 1/6 of its value on earth. Suppose he hit the ball with a speed of 35 m/s at an angle 27 ∘ above the horizontal. Part A How long was the ball in flight? Express your answer using two significant figures.

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Mercedes Cobb · Answer 1

On the moon, the gravitational acceleration is 1/6 of 9.8 m/s², so

g = 9.8/6 = 1.633 m/s²

Launch speed = 35 m/s

Launch angle = 27° above the horizontal.

Therefore,

The horizontal velocity is

u = 35*cos (27) = 31.1852 m/s

The vertical launch velocity is

v = 35*sin (27) = 15.8897 m/s

Part A

When the ball reaches maximum height, the time requires is given by

0 = v - gt

t = v/g = 15.8897/1.6333 = 9.7286 s

This is one half of the time of flight, which is

2*9.7286 = 19.457 s

Answer: 19.46 s (2 sig. figs)