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.

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

19.46 s (2 sig. figs)

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