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Homework Help: Finding force

  1. Nov 16, 2007 #1
    1. The problem statement, all variables and given/known data

    An 88 kg man drops from rest on a diving board 3.3m above the surface of the water and comes to rest 0.58 seconds after reaching the water.

    Acceleration due to gravity is 9.81 m/s^s

    What force does the water exert on the man? in N


    2. Relevant equations

    F = ma
    a = v/t

    3. The attempt at a solution

    3.3/0.58 = 5.689655172 m/s = v

    F = 88 * 5.689655172 = 500.6896552 N

    F = 500.6896552 N

    Did I do this right?
     
  2. jcsd
  3. Nov 16, 2007 #2

    mgb_phys

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    First you need the speed he hit the water - you got this wrong
    v^2 = u^2 + 2 a h
    V^2 = 2 * g * 3.3 , v = 8.05m/s

    Then the accelaration (rate of slowing down) in the water
    V = u + a t,
    so 0 = 8.05 - a * 0.58
    a = 13.8 m/s^2

    Then F = ma
     
  4. Nov 16, 2007 #3
    Don't we need to know the force the diving board exerts on the man, too? http://www.turboconnection.com/pics/smilies/hmm.gif [Broken]
     
    Last edited by a moderator: May 3, 2017
  5. Nov 16, 2007 #4
    Never mind. He doesn't drop on to the diving board; he drops off the diving board and into the water.
     
  6. Nov 16, 2007 #5
    Start with this equation to figure the speed he's at when he hits the water:

    (1) [tex]a \Delta{x} = \frac{1}{2}(v^2-v_0^2)[/tex]

    Then use this equation to find his acceleraction (deceleration) when he's in the water:

    (2) [tex]x=x_0+v t+\frac{1}{2}a t^2[/tex]

    Equation 2 can be simplified into terms of only velocity and acceleration by differentiating it with respect to time, as such:

    (3) [tex]v=v_0+at[/tex]

    Then apply the definition of force:

    (4) [tex]F=ma[/tex]

    That should do it.
     
  7. Nov 16, 2007 #6
    Wouldn't a = 4.667 because 8.05 * 0.58 = a

    never mind 8.05 / .58 = 13.873
     
    Last edited: Nov 16, 2007
  8. Nov 16, 2007 #7
    How do I find the velocity?
     
  9. Nov 16, 2007 #8
    yay i got it right! thanks a lot!
     
  10. Nov 16, 2007 #9
    Use the first equation. You know [tex]a[/tex]:

    [tex]a=-g=-9.8\frac{m}{s^2}[/tex]

    You know initial velocity:

    [tex]v_0=0[/tex]

    And you know the change in distance:

    [tex]\Delta{x}=-3.3m[/tex]

    Actually, it should be

    [tex]\Delta{y}=-3.3m[/tex]

    since we're vertical.

    Solve for [tex]v[/tex].
     
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