anonymous
  • anonymous
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?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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Astrophysics
  • Astrophysics
Write down all the variables which are present in this problem
Astrophysics
  • Astrophysics
Think about kinematics for now, what does it mean maximum heigh h?
anonymous
  • anonymous
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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Astrophysics
  • Astrophysics
at maximum height vf = 0
Astrophysics
  • Astrophysics
Yes, you can use the ratio to find the height on mars
anonymous
  • anonymous
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?
anonymous
  • anonymous
I'm mostly lost on which formula is the most appropriate to use for this question.
Astrophysics
  • Astrophysics
Was your first question in terms of h?
anonymous
  • anonymous
Yes it was.
Astrophysics
  • Astrophysics
So for that you took the ratio of the kinematic equation v^2 = v0^2+2ah then you should have 2.64H
anonymous
  • anonymous
Yes that is what I did!
Astrophysics
  • Astrophysics
Ok, so now use since you want time, what kinematic equation can we use?
anonymous
  • anonymous
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?
Astrophysics
  • Astrophysics
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 \[0=v_0t+1/2at^2\]
Astrophysics
  • Astrophysics
The reason it is 0 is because it goes up and then it returns to the ground
anonymous
  • anonymous
I see! Thanks. Now we need to find V0 to solve for t? Setting V0 to 0 only makes t = 0.
Astrophysics
  • Astrophysics
You should end up with \[(at)_E=(at)_M\] where and E and M are the planets
Astrophysics
  • Astrophysics
You should note the answer is similar to a but the quantity of course is different
anonymous
  • anonymous
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!
Astrophysics
  • Astrophysics
Sounds good! And np :)

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