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One-Dimensional Kinematics Problem

  1. Sep 5, 2006 #1
    Hey guys,

    There is one more question I am having a problem with. Here it is:

    You jump from the top of a boulder to the ground 2.07 m below. Calculate your deceleration on landing. Assume that in order to soften your landing, your legs will bend 0.510 m. Give your answer in units of g. Do not enter unit.

    Here is what I have done so far. I have used the height to find out how fast you are when you reach the ground. I got v= 6.37 m/s. After that I know I have to use the distance of bending to determine over what distance you decelerate. I know I use the formula that relates the velocity to the distance. So I should use v^2 = v_o^2 - 2g(xf-xi). THat is the same formula I used to get the first answer. Maybe I am using the wrong formula. This problem and the one I posted before I have made like 10 attempts on LON CAPA and I don't know what to do anymore. Please help. Thank you,


    You can also send me an email if you dont want to post on the board.
  2. jcsd
  3. Sep 5, 2006 #2
    You correctly calculated how fast you are when you reach the ground:
    v2 = 2gh

    Relating work, Newton 2nd Law and Work-Energy Theorem we have:
    W = max = \Delta KE = 0 - 0.5mv2
    max = 0 - 0.5mv2
    max = -0.5m2gh
    max = -mgh
  4. Sep 5, 2006 #3
    I don't know any of those formulas or anything you listed above. I just started physics so we haven't gotten to those yet. Thanks for letting me know I got the first part right. I need things in simple terms I guess.
  5. Sep 5, 2006 #4


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    That equation is almost correct, you should however be using the general case (not just for objects falling under gravity) thus;

    [tex]v^2 = v_{0}^{2} + 2a(x_{f} - x_{i})[/tex]

    Where a is acceleration. Don't forget to divide your answer by 'g' to obtain your answer in terms of g.
  6. Sep 5, 2006 #5
    so v^2 is 40.61, what should I put in for v_o? and what should I put in for the change in x in that equation. If i put in 0 as initial velocity and change in x as 0.510 i end up with acceleration -39.81, divide that by 9.81 and i get -4.06 which was incorrect according to lon capa.
  7. Sep 5, 2006 #6


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    Note quite, v2 represents the final velocity, which in this case is zero. V0 would be the velocity at which he reaches the ground. Do you follow?
  8. Sep 5, 2006 #7
    even if i switch the two v's i end up with the same answer just different sign, unless i am not supposed to put in 0.510 for x. I am ready to quit physics.
  9. Sep 5, 2006 #8


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    Well, I've quickly checked your calculations and I can find nothing wrong. Have you tried entering the modulus of the acceleration (without the negative sign). Make sure your not entering any units as the value has no units. Are you using their quoted value of g?
  10. Sep 5, 2006 #9
    Hey, thank you so much for that advice Hoot. I was putting in -4.06 and i tried it without the negative and it worked. THANK YOU. Can you look at my other problem I posted ??? please!
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