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Homework Help: Physics HW- Falling Bodies, Part 3

  1. Nov 6, 2013 #1
    1. The problem statement, all variables and given/known data

    The moon is smaller than the earth and has a weaker gravity. On the moon's surface, the acceleration due to gravity is 1.6 m/s^2. You are on the moon and you throw a moon rock straight up. It reaches a height of 44 m.

    a) Find the moon rock's initial velocity.

    b)Find the total time of the moon rock's flight.

    2. Relevant equations
    v0^2=v^2-2αΔy
    t^2 =2y/a



    3. The attempt at a solution

    a) v0^2 =0-2(1.6 m/s^2)(44 m)
    √v0^2=√140.8

    v0=11.9 m/s



    b)
    t^2 =2y/a
    t^2= 88 m/1.6 m/s^2

    Taking the square roots of both sides:
    t=55 s.

    Are these answers correct? This is a newer concept to me, so input would be greatly appreciated!
     
  2. jcsd
  3. Nov 6, 2013 #2

    CAF123

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    You have neglected the fact that the rock had an initial velocity. Also, you have missed a negative in your calculation of vo. As you wrote it, you would have the sqrt of a negative.
     
  4. Nov 6, 2013 #3
    Okay, thank you! Was first answer was correct, besides the missed negative? I thought that the two negatives canceled out, but I probably did that step wrong.

    Did I use the wrong formula? I can't seem to find a better one on my equation sheet. Do you have a suggestion?

    Thank you!!!
     
  5. Nov 6, 2013 #4

    CAF123

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    The important thing to note is that you are dealing with vector quantities. The kinematic equation you used was ##v^2 = v_o^2 + 2as##. ##v = 0## at the apex of the throw. Set up a coordinate system with which to choose the positive direction for each vector quantity. Taking upwards as positive, you see that a = -g since gravity acts downwards. vo and s are positive since the ball is thrown upwards. Hence you get ##-v_o^2 = -2gs## and now you see the negatives cancel. If you define positive as downwards, this is also fine.
    Not necessarily the wrong kinematic equation, but you are assuming the initial velocity is zero.
     
  6. Nov 6, 2013 #5
    I understand what you're saying, besides the part about the negative. Are you agreeing that the negatives cancel out, but I stated something wrong in my formula?

    Should it look something like this?

    v0^2=v^2-2aΔy
    v0^2=0-2(-1.6 m/s^2) (44 m)
    -v0^2= -√140.8
    =11.9 m/s

    So would it be the same answer, but the negatives cancel differently earlier?


    Using the formula t^2=2y/a , I can't seem to figure out where the inital velocity comes into play, although it has to come in somewhere!

    Thanks again!
     
  7. Nov 6, 2013 #6

    CAF123

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    vo2=v2-2aΔy is the same as v2=vo2+2aΔy, just rearranged. What you have done is completely correct, but I don't see why you took the negative onto the LHS. The only reason I had a negative on the RHS in my previous post is because I started with v^2 = v_o^2 + 2as and rearranged for v_o^2.

    I believe you used the right kinematic equation in one of your other posts: it is ##\Delta y = v_o t + \frac{1}{2}at^2##. If the initial velocity is zero, this reduces to ##t^2 = 2 \Delta y/a## as you have been using. In this question, use the general version.
     
  8. Nov 6, 2013 #7
    Oh! I'm not sure what LHS is, but it was most likely an odd accident. Thanks!


    Thank you! Now I get it!

    Δy=[v][0]t+1/2 a[t][2]
    Δy=11.9 m/s(t) +1/2 [1.6(t)][2]

    So what's next, since the inital velocity ISN'T zero? Do I divide both sides by t?
     
  9. Nov 7, 2013 #8
    Oh, I forgot I have Δy.

    It should look like this:

    44 m=11.9 m/s(t) +1/2 [1.6(t)][2]

    So now I have everything but the time, so all I really have to do is simplify?

    Thanks again!
     
  10. Nov 7, 2013 #9

    CAF123

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    I would say so, but using a different kinematic eqn I obtain a different t, which is puzzling me. Perhaps another user could clarify.
     
  11. Nov 7, 2013 #10
    Hmm. That's odd. I wish I could help clarify, but I only know a few basic equations at this time. Do you think it would help if I tried attaching my formula sheet?
     
  12. Nov 7, 2013 #11

    CAF123

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    When you solve your eqn, you will obtain two values of t since it is quadratic. If you average those values of t, you will obtain exactly the answer I get using another kinematic eqn. But I don't see the need to average and why there exists two times when the ball is at a height 44m. Let's wait for another user to clarify.
     
  13. Nov 7, 2013 #12
    Okay, thank you!
     
  14. Nov 7, 2013 #13

    Doc Al

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    Careful with signs. The acceleration should be -1.6 m/s^2.

    The real problem is that this equation is very sensitive to the exact initial velocity you use. Instead of using 11.9, use a more "exact" answer from your first part. Then the two solutions will converge to be the same.

    Even better: Use distance = ave velocity * time. What's the average velocity?
     
  15. Nov 7, 2013 #14
    Okay, thanks!

    vavg=Δx/Δt


    Only I'm not sure of the average velocity. Should I use the previous equation and average my answers?
     
  16. Nov 8, 2013 #15

    Doc Al

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    No. For uniformly accelerated motion you can find the average velocity using this:
    Vave = (V0 + Vf)/2

    You already found the initial velocity earlier and you know the final velocity.
     
  17. Nov 8, 2013 #16
    Thanks! I can't believe I forgot that equation.

    Something like this?

    vavg= (11.9 m/s+0)/2≈ 5.95 m/s
     
  18. Nov 8, 2013 #17

    Doc Al

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    Exactly. Now you can solve for the time to rise without worrying about any quadratics.

    I just realized that they ask for the total time, not just the time to rise. Be careful!

    You can use a quadratic equation to solve for the total time: It's easy!
     
  19. Nov 8, 2013 #18
    Okay, thank you! Do I use this equation next? vavg=Δx/Δt
     
  20. Nov 8, 2013 #19

    CAF123

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    Just for completeness, to find the time taken to rise in one equation and without any quadratics consider using v = v0 + at.
     
  21. Nov 8, 2013 #20
    Just to clarify, is this a formula I need to apply to this problem, or are you giving me a helpful formula for future reference? Thank you!
     
  22. Nov 8, 2013 #21

    CAF123

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    It is just another of the standard kinematic equations. I suggested it is an alternate way to find the time taken for the rock to rise to its apex. You know v, v0 and g so you can find t - try it and see that it is the same as you get using the previous methods.
     
  23. Nov 8, 2013 #22
    Thank you! I will definitely keep this in mind. I don't have time to try it right now, because I am working on another three other physics problems, but it will prove useful very soon. Do you know the next step?
     
  24. Nov 8, 2013 #23

    CAF123

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    There are multiple ways to find the time taken for the rock to fall back down: Two are;

    What is the rock's velocity at the top of the throw? (Let this be v0)
    How far does it fall? (Let this be Δy)
    Now using one of the kinematic equations you can find v. Then use the kinematic equation I suggested in my last post to find t. (Be careful with signs)

    or

    Have you covered conservation of energy yet?
     
  25. Nov 8, 2013 #24
    Thank you! Are we using v=v0+at to find v?
     
  26. Nov 8, 2013 #25

    CAF123

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    Provided you have all the correct data, you can use whatever equation you like - in this case, to find v from that eqn you would need to know the time taken for the rock to fall to the ground, but this is what you are looking for. Find instead a kinematic equation relating v0, Δy and v so you can find v.

    Once you know v, then use v = v0+at to find t.

    The time taken for the rock to fall back down is the same time it would take if somebody just dropped it from this height. This is because the rock's velocity is zero at the apex.
     
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