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Jumping off a boulder

  1. Aug 19, 2012 #1
    1. You jump from the top of a boulder to the ground 1.5m below. Estimate your deceleration on landing.

    2. v = v0 + at, x = x0 + v0t + 0.5at^2, v^2 = v0^2 + 2a(x-x0)

    3. I initially thought that it was gravity (-9.8 m/s) but it's not the answer. I also tried making v into 0 and solving for v0 but it ends up wrong too.

    I don't understand why it isn't just gravity..is this not a free fall? I'm going crazy..please help!
     
  2. jcsd
  3. Aug 19, 2012 #2
    9.8m/s^2 seems right though. What was the answer given?
     
  4. Aug 19, 2012 #3
    Gravity is your acceleration during the fall. This problem is asking for your deceleration when you land. To answer that, you need to know two things: how fast were you going when you hit, and how long did it take you to stop?
     
  5. Aug 19, 2012 #4

    CAF123

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    Free fall while falling, but the question is about on landing. When you land, you experience a large acceleration upwards
    EDIT: Aimless made a similar comment to mine
     
  6. Aug 19, 2012 #5
    How do you solve when you don't know v0 and t? Would you still set x to 1.5?
     
  7. Aug 19, 2012 #6
    The answer given is -29m/s^2.
     
  8. Aug 19, 2012 #7
    Why don't you know v0? What is v0? How would you go about calculating it?

    Why don't you know t? Can you estimate it?
     
  9. Aug 19, 2012 #8

    CAF123

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    Set [itex] x-x_o = -1.5, [/itex] if you define downwards as negative.
     
  10. Aug 19, 2012 #9
    .....I have no idea..........I guess you don't know v0 because you are jumping?
     
  11. Aug 19, 2012 #10
    You miss the point of my question. v0 is a velocity, but where are you measuring that velocity? What is the physical meaning of v0?

    Physics problems are all about how you translate from the description of the problem into the mathematical quantities you are trying to solve, so before you start plugging things into equations you have to get straight in your head exactly what the physical meaning of each quantity is.

    In this case, we are trying to calculate a deceleration. Our final velocity is after landing, so what is it? In order to decelerate, we must have had some velocity to begin with. Where did that velocity come from?
     
  12. Aug 19, 2012 #11
    I guess the v0 would be the velocity during the falling..so how fast the person is falling and the final velocity would be 0 because the person has landed.
     
  13. Aug 19, 2012 #12
    Almost, but not quite. v0 is actually the velocity at the end of the fall, just before impact (just prior to the start of the deceleration). Given that, how would you go about computing v0?

    You are correct that the final velocity is 0.
     
  14. Aug 19, 2012 #13
    I guess you could divide 1.5 by the time it takes to reach the end of the fall...?
     
  15. Aug 19, 2012 #14

    CAF123

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    Dividing 1.5 by the time to reach the ground will give you the persons average velocity. You want the velocity prior to the impact with the ground.
    You can determine this by energy conservation or by using the kinematic relations.
     
  16. Aug 19, 2012 #15
    If we estimate that your height will be less 0.5m(1/4 of estimate height of 2m) when finally stop.
    We assume this value where you have to bend to soften the landing.
    We use this value as distance travelled in deceleration motion.

    v2=2g(1.5)
    For deceleration
    -2g(1.5)=2a(0.5)

    Estimate
    a=-29m/s2
     
  17. Aug 19, 2012 #16

    CAF123

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    A typical bend in your legs when you touch the ground is about the 0.5m mark that azizlwl mentioned. Using either energy conservation or kinematics to determine the velocity before impact gives [itex]v_o, [/itex] you know [itex] v_f, [/itex] and [itex] s [/itex] here is the 0.5m. Solving for a will yield an estimated -29m/s2.
     
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