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 1/6 of its value on earth. Suppose he hit the ball with a speed of 27 m/s at an angle 36 degree above the horizontal. How long was the ball in flight? Express your answer to two significant figures and include the appropriate units. How far did travel? Express your answer to two significant figures and include the appropriate units. 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.

Explanation / Answer

Given,

v = 27 m/s ; theta = 36 deg

A)the time for which its in air will be given by:

t = 2 v0y/g (deduced from s = ut + 1/2 at^2)

t = 2 x 27 x sin36/(9.8/6) = 19.43 s

Hence, t = 19.43 s

b)the distance will be:

x = v0^2 sin(2 theta)/g

x = 27^2 sin(2x36)/9.8/6 = 424.48 m

Hence, x = 424.48 m

c)on earth, using the same relation

xe = 27^2 sin(2x36)/9.8 = 70.75 m

Hence, xe = 70.75 m

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