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Two dimensional motion.

  1. Nov 10, 2011 #1
    1. The problem statement, all variables and given/known data
    A stuntman must jump in a pool from a 50.0 meter tall building. The pool is a horizontal distance of 21.0 meters from the base of the building.

    a) What is the initial velocity of the stuntman to be able to land in the pool?
    b) Do you think this velocity is reasonable for a human?
    c) Realistically, how tall must this building be for the stuntman to be able to make the landing safely?


    2. Relevant equations
    Vf^2 = Vi^2 + 2aΔx
    Vf = Vi + aΔt
    Xf-Xi = Δx
    Given:
    Xix = 0m
    Xfx = 21m
    Xiy = 50m
    Xfy = 0m
    Δx = 21m
    Δy = -50m
    a= -9.81m/s^2
    Viγ = 0m/s

    Relevant Unknowns:
    Δt = ?
    Vfγ = ?

    3. The attempt at a solution
    Vf^2 = Vi^2 + 2aΔx
    Vf^2 = 0^2 + 2(-9.81m/s^2)(-50m)
    Vf^2 = 2(9.81m/s^2)(50m)
    Vf^2 = 981
    Vf = 31.3 m/s

    Vf = Vi + aΔt
    31.3 = 0 + (-9.81m/s^2) Δt
    31.3 = -9.81m/s^2 (Δt)
    31.3 = Δt
    -9.81m/s^2

    Yeah, you see the issue, I made a stupid mistake and my brain is on 0 hours of sleep. and I have school in 1 hour.
     
  2. jcsd
  3. Nov 10, 2011 #2

    BruceW

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    Homework Helper

    You were close, Vf^2 = 981 is right. But then you chose the positive root. The actual answer is the negative root Vf = -31.3 because the stuntman is falling downwards. So the change in time is the positive value of the answer you got.
     
  4. Nov 10, 2011 #3
    But mathematically that would take me into irrational numbers, E.g. 31.3i

    Which would mean I would have to do some algebraic things I don't even remember how to do anymore.

    I just realized that square roots have a + or -, never mind.

    See? My brain doesn't have it's sensitivity to serotonin at the moment.
     
  5. Nov 10, 2011 #4

    BruceW

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    Homework Helper

    No worries, good luck finishing the question.
     
  6. Nov 10, 2011 #5
    Alright, this time I did it right.

    Garcias
     
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