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 acceleration due to gravity on the moon is 1/6 of its value on earth. Suppose he hits the ball with a speed of 16 m/s at an angle 15 degree above the horizontal. How much farther did the ball travel on the moon than it would have on earth? Express your answer using two significant figures. Delta L = m For how much more time was the ball in flight on the moon than it would have on earth? Express your answer using two significant figures. Delta t = s

Explanation / Answer

PROJECTILE

along vertical

voy = vo*sintheta

acceleration ay = -g

after falling on ground

displacement Y-Yo = 0

from equation of motion

y-y0 = voy*T + 0.5*ay*T^2

0 = v0y*T - (1/2)*g*T^2

T = 2*v0y/g = 2*v0*sintheta/g

along horizontal

acceleration ax = 0

inital velocity vox = v0*costheta

displacement x-x0 = L

from eqation of motion

x = v0x*t + (1/2)*ax*t^2

L = v0*costheta*2*v0*sintheta/g

L = v0^2*sin(2theta)/g

part(A)

on earth

g = 9.8

L1 = 16^2*sin(2*15)/9.8 = 13 m

on moon

g = 9.8/6

L2 = 16^2*sin(2*15)/(9.8/6) = 78 m

dL = L2 - L1 = 65 m   <<<==========ANSWER

====================

part B

T = 2*vo*sintheta/g

on earth

T1 = 2*16*sin15/9.8

T1 = 0.84 s

on moon

T2 = 2*16*sin15/(9.8/6) = 5.1 s

dt = T2 - T1 = 5.1 - 0.84 = 4.3 s <<<<==========ANSWER

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