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 24 m/s at an angle 55 above the horizontal.

a) 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.

b) For how much more time was the ball in flight?

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

Explanation / Answer

a)

initial speed , v = 24 m/s

theta = 55 degree

Range of ball on earth , R = v^2 * sin(2 * theta)/g

R = 24^2 * sin(2 * 55 )/9.8

R = 55.23 m

Now , on moon , as gmoon = 1/6 * gearth

hence , Rmoon = 6 * Rearth

extra distance on moon = 55.23 * (6 - 1)

extra distance on moon = 276.15 m

the extra distance on moon is 276.15 m

b)

time of flight on moon ,

tfm = 2 * v * sin(theta)/gmoon

tfm= 2 * 24 * sin(55)/( 9.8/6)

tfm = 24.1 s

Now , time of flight on earth = 24.1/6 = 4.01 s

more time = 24.1 - 4.01

more time = 20.1 s

the more time of flight on moon is 20.1 s

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