Gravity Strength: Moon vs Earth Ratio of Height Reached

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The strength of gravity on the moon is one-sixth that of Earth, affecting the height a ball can reach when thrown upward. The ratio of the height reached on the moon to that on Earth, given the same initial velocity, is determined to be 6. This conclusion is drawn from kinematic equations that show the distance traveled is proportional to the inverse of gravity, leading to a consistent ratio regardless of initial velocity. The calculations confirm that the ratio does not depend on the initial velocity, reinforcing the correctness of the result. Thus, the ratio of heights is indeed 6.
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Homework Statement



The strength of gravity on the moon is 1/6 of that on earth. If a ball is thrown straight upward on the moon with initial velocity Vo the ratio of the height it reaches to the height it would reach on earth, if also thrown upward with velocity Vo is:

2
1/6
6
depends on Vo
36

Homework Equations


The Attempt at a Solution


EDIT: I got 6 I want to know if this is correct because it is a review question for an exam I have in an hour. Can someone help me out? I solved 2 kinematic equations with Vo with the only knowns being the gravity acclerations and got x on Earth = Vo^2 / (2g) and x on the moon to be = 6Vo^2 / (2g)
 
Last edited:
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While the distances the ball travels upward on either the Earth or the moon will depend on V_o, the ratio of them will not.

Try to set up an equation of motion for each situation. Using the two equations you should see that you can find an equation for the ratio that is not dependent on V_o.
 
right I did an edit to show I got 6 is that correct?
 
Correct.:smile:
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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