A certain volcano on earth can eject rocks vertically to a maximum height H. The acceleration due to gravity on Mars is 3.71 m/s2, and you can neglect air resistance on both planets. If the rocks are in the air for a time T on earth, for how long (in terms of T) will they be in the air on Mars?

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A certain volcano on earth can eject rocks vertically to a maximum height H. The acceleration due to gravity on Mars is 3.71 m/s2, and you can neglect air resistance on both planets. If the rocks are in the air for a time T on earth, for how long (in terms of T) will they be in the air on Mars?

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Write down all the variables which are present in this problem
Think about kinematics for now, what does it mean maximum heigh h?
So there is a previous part to this question which asks for the ratio between maximum height a rock will reach on Mars and that of a rock on Earth. \[\frac{ Hm }{ H } = 2.64\]in which Hm is the max height a rock on Mars will travel and H is that of a rock on Earth. I know that the initial velocity is 0 and I manipulated a kinematics equation to solve for that ratio. But what should I do to get time? Should I do the same steps incorporating 2.64 somewhere? I can tell you how I solved for 2.64.

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Other answers:

at maximum height vf = 0
Yes, you can use the ratio to find the height on mars
What is my firsrt step? Do I use the formula vf^2 = vi^2 + 2ad and substitute the ratio into d and also set vf = 0?
I'm mostly lost on which formula is the most appropriate to use for this question.
Was your first question in terms of h?
Yes it was.
So for that you took the ratio of the kinematic equation v^2 = v0^2+2ah then you should have 2.64H
Yes that is what I did!
Ok, so now use since you want time, what kinematic equation can we use?
d = Vi(t) + (1/2)a(t)^2 since we know the acceleration on earth, Vi = 0 and we use 2.64 somehow for the displacement?
Yes we use that equation, specifically \[y-y_0=v_0t+1/2at^2\] the direction would be the y direction but I'm not to worries about putting subscripts, so the y-y0/ d = 0 so then we will be left with \[0=v_0t+1/2at^2\]
The reason it is 0 is because it goes up and then it returns to the ground
I see! Thanks. Now we need to find V0 to solve for t? Setting V0 to 0 only makes t = 0.
You should end up with \[(at)_E=(at)_M\] where and E and M are the planets
You should note the answer is similar to a but the quantity of course is different
So the answer was actually the same, but the method to get there was different. Thank you so much and thank you for being patient for me!
Sounds good! And np :)

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