A golf ball being hit on the moon vs Earth

AI Thread Summary
The discussion focuses on calculating the range of a golf ball hit on the moon and Earth, using the range equation R=(Vi^2sin2theta)/g. The user calculated the gravitational acceleration on the moon as 1.63 m/s² and found the golf ball would travel 452 meters on the moon and 75.2 meters on Earth. The use of the range equation is confirmed as appropriate since the projectile lands at the same height from which it was launched. The conversation also notes that the absence of atmospheric drag on the moon simplifies the calculations, although it doesn't account for potential aerodynamic effects from the ball's design. Overall, the calculations and approach are validated, emphasizing the differences in range due to varying gravitational forces.
Jregan
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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?
Relevant 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
 
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Jregan said:
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.
 
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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...
 
Thank you!
 

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