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A golf ball being hit on the moon vs Earth

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Homework Statement
A golf ball is hit on the moon which has a free-fall acceleration 1/6 of its value on earth. The ball is hit at a speed of 28m/s at an angle of 35 degrees. How far did the golf ball travel? Ignoring air resistance, how far would it travel on earth?
Homework Equations
R=Vi^2sin2theta/g
I am just not sure if I did this properly. My professor hasn't really gone over when to use the range equation but I would assume range would equal the distance traveled therefore can be used for this problem. If not the how would I go about solving this?

I did 1/6*9.8=1.63 for g on the moon

Then used R=(Vi^2sin2theta)/g

so on the moon
R=((28^2)sin2*35)/1.63=452m

and on earth
R=((28^2)sin2*35)/9.8=75.2m
 

haruspex

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Homework Statement: A golf ball is hit on the moon which has a free-fall acceleration 1/6 of its value on earth. The ball is hit at a speed of 28m/s at an angle of 35 degrees. How far did the golf ball travel? Ignoring air resistance, how far would it travel on earth?
Homework Equations: R=Vi^2sin2theta/g

I am just not sure if I did this properly. My professor hasn't really gone over when to use the range equation but I would assume range would equal the distance traveled therefore can be used for this problem. If not the how would I go about solving this?

I did 1/6*9.8=1.63 for g on the moon

Then used R=(Vi^2sin2theta)/g

so on the moon
R=((28^2)sin2*35)/1.63=452m

and on earth
R=((28^2)sin2*35)/9.8=75.2m
Looks good.
 

kuruman

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The use of the range equation is appropriate in this case. You can use it whenever the projectile lands at the same height from which it was launched.
 
Last edited:
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Simpler calculation without the atmosphere, no drag, but also no allowance required for 'spin' which, with the dimples, may provide some aerodynamic lift and extend range thus.

IIRC, one of the braw Apollo moon-walkers took a golf-ball or two as part of his 'personal allowance', whacked them 'into the rough' with an adapted sampling tool...
 
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Thank you!
 

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