A golf ball being hit on the moon vs Earth

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Homework Help Overview

The discussion revolves around the physics of projectile motion, specifically comparing the distance a golf ball travels when hit on the moon versus on Earth. The problem involves calculating the range of the golf ball given its initial speed and launch angle, while considering the different gravitational accelerations on the two celestial bodies.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the range equation for projectile motion, questioning its appropriateness for this scenario. They calculate the gravitational acceleration on the moon and use it to find the range on both the moon and Earth. Some participants affirm the use of the range equation, while others provide context about the conditions affecting projectile motion.

Discussion Status

The discussion is active, with participants validating the use of the range equation and exploring the implications of different gravitational forces. There is acknowledgment of the simplifications made in the calculations, such as ignoring air resistance and other factors that could affect the range.

Contextual Notes

Participants note the absence of air resistance in the calculations and discuss the potential effects of factors like spin and atmospheric conditions on the golf ball's trajectory.

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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