Finding the max range of a jump on the moon using kinematics

AI Thread Summary
To determine the maximum jump range on the Moon, the initial jump speed on Earth was calculated to be approximately 6.26 m/s. Using this speed and the Moon's lower gravity of 1.6 m/s², the maximum jump distance was found to be about 24.03 m. The discussion highlighted that the range is inversely proportional to the gravitational acceleration, illustrating that as gravity decreases, the jump distance increases significantly. This relationship simplifies calculations, avoiding complex arithmetic. Overall, the findings emphasize the impact of reduced gravity on jump distances in different celestial environments.
garcia1
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Homework Statement


A person can jump a maximum horizontal
distance (by using a 45 ◦
projectile angle) of
4 m on Earth.
The acceleration of gravity is 9.8 m/s
2
.
What would be his maximum range on the
Moon, where the free-fall acceleration is g
6 ?
Answer in units of m.



Homework Equations



Kinematics equations

The Attempt at a Solution



All I can think to do is set kinematics equations for the x and y-axis equal to each other by another variable, such as time. This is as far as I can figure out with this problem.
 
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hi garcia1! :smile:

(have a degree: ° and try using the X2 icon just above the Reply box :wink:)

from the first part, find the speed at which this person can jump

then use that speed with the Moon's g to find how far he will jump on the Moon …

what do you get? :smile:
 
24.0304m/s, by using the fact that vf = 0, and then solving for VoY and getting 6.26m/s. I then plugged this into:

Y = Vy^2 - VoY^2 / (2*-9.81m/s^2 / 6) = 12.01518

Multiplying by two because I used Vf = 0 at the top of the jump, I got 24.0304m. It was right!
 
:biggrin: Woohoo! :biggrin:

ok, now have you noticed that the range is inversely proportional to the gravity (24/4 = 6)?

can you prove that, and so avoid all the tedious arithmetic? :wink:
 
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