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anonymous

  • one year ago

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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  1. Astrophysics
    • one year ago
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    Write down all the variables which are present in this problem

  2. Astrophysics
    • one year ago
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    Think about kinematics for now, what does it mean maximum heigh h?

  3. anonymous
    • one year ago
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    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.

  4. Astrophysics
    • one year ago
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    at maximum height vf = 0

  5. Astrophysics
    • one year ago
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    Yes, you can use the ratio to find the height on mars

  6. anonymous
    • one year ago
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    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?

  7. anonymous
    • one year ago
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    I'm mostly lost on which formula is the most appropriate to use for this question.

  8. Astrophysics
    • one year ago
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    Was your first question in terms of h?

  9. anonymous
    • one year ago
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    Yes it was.

  10. Astrophysics
    • one year ago
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    So for that you took the ratio of the kinematic equation v^2 = v0^2+2ah then you should have 2.64H

  11. anonymous
    • one year ago
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    Yes that is what I did!

  12. Astrophysics
    • one year ago
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    Ok, so now use since you want time, what kinematic equation can we use?

  13. anonymous
    • one year ago
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    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?

  14. Astrophysics
    • one year ago
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    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 \[0=v_0t+1/2at^2\]

  15. Astrophysics
    • one year ago
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    The reason it is 0 is because it goes up and then it returns to the ground

  16. anonymous
    • one year ago
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    I see! Thanks. Now we need to find V0 to solve for t? Setting V0 to 0 only makes t = 0.

  17. Astrophysics
    • one year ago
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    You should end up with \[(at)_E=(at)_M\] where and E and M are the planets

  18. Astrophysics
    • one year ago
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    You should note the answer is similar to a but the quantity of course is different

  19. anonymous
    • one year ago
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    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!

  20. Astrophysics
    • one year ago
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    Sounds good! And np :)

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