Solving the Vertical Throwing Object on Moon vs Earth Gravity

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An object thrown vertically on the Moon will reach a height six times greater than on Earth due to the Moon's gravity being one-sixth that of Earth's. To calculate the height, kinematic formulas are necessary, particularly those involving initial velocity and acceleration. A suggested initial velocity for calculations is 9.8 m/s. The discussion emphasizes the importance of understanding the relationship between acceleration and distance in vertical motion. Participants are encouraged to find relevant formulas and apply them to demonstrate their calculations.
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If an object is thrown vertically upward on the moon how many times higher will it go than it would on Earth if they both have the same initial vel.

Acceleration on moon is 1/6 of what it is on earth

I need some help on solving this. Thank You
 
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Pick an initial velocity, let's say 9.8 m/s to keep it simple. On Earth do you know how high it would go?
 
how do you calculate to get the height?
 
Do you have kinematics formulas? Is this a high school class?
 
yes but i don't have the formulas or the textbook with me but only the questions because this is homework over the summer
 
Without doing any math, do you have a guess as to the answer?
 
is it 96.04m?
 
no. but I didn't mean the answer to my question in the 2nd post of this thread. I meant can you guess at the answer to your original question?
 
it'll go six times higher?
 
  • #10
That wasn't too hard, was it? :)
 
  • #11
o really? that's the right answer but can u teach me how to do it mathematically? because i have to show my work and steps in how to get the answer
 
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  • #12
Without doing your work for you, I can't. But there's a few threads similar to this one posted today on this same forum that will give you all the formulas you need. Get the formula, and determine why distance is linear.
 
  • #13
can someone provide me the formulas i can't seem to find any of them in this forum?
 
  • #14
If the acceleration is -a (negative because it is downward) and there is an initial velocity v0 upward, the velocity at time t is v= v0- at and the height is v0t- (1/2)at2. Can you figure out from that the time until the highest point and that height, both in terms of a and v0?
 
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