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Homework Help: Motion in a straight line

  1. Sep 14, 2009 #1
    1. The problem statement, all variables and given/known data

    Two divers jump from a 2.70meter platform. One jumps upward at 1.60meters/second and the 2nd steps off the platform as the first passes it on the way down. What are their speeds as they hit the water? Which hits the water first and by how much?

    2. Relevant equations
    Vf^2= Vi^2 + 2ad (Vf being final velocity and Vi being initial velocity)
    Y=Y0 + ViT - 0.5gt (Y and Ynaught being the distance in the y-axis direction, T being time, and g being gravity)

    3. The attempt at a solution

    So i used the equation Vf^2=Vi^2 + 2ad to find the speed at which the diver that walked off the platform hit the water:

    Vf^2= 0 + 2(-9.8m/s^2)(0-2.7m) = sqroot[(-18.6m/s^2)(-2.70m)]= 7.09m/s

    I'm somewhat sure that to find the speed of the other diver I have to use that 2nd equation that I provided and 1 more additional equation, but yeah, Im stuck. Can anyone help? Equations and explanations are much appreciated, thanks

    Oh, and off the bat I'm going to say the diver that jumped from the platform will reach the water first as opposed to the diver that walked off. But I don't know how to show that, so yeah...
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Sep 14, 2009 #2


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    Using the second equation, find the time taken by both of them.
    For both of them, displacement is the same. Initial velocity for walker is zero.
  4. Sep 14, 2009 #3
    Alright, so for the diver that jumps off, I used the equation Y=Y0 + Vit -0.5gt^2. Using some algebraic manipulation, t= [Vi +-sqroot(Vi^2 +2Y0g)]/g

    1.6m/s +- sqroot((1.6^2)+2(2.7m)(9.8m/s^2))/9.8m/s^2. I get 2 answers, one of them is t=0.92s and the other is t=-0.59s, so im rejecting the negative and accepting 7=0.92s as the time for the jumping diver.

    For the diver that walks off, I used the same equation, except i made Vi=0. and I got 0.74s as my positive time.

    So I'm guessing this proves the jumping diver gets to the water before the diver that walks off due to short time right?

    And what about the speed for both of them hitting the water? Was my first equation right in doing that?

    let me know if i did something wrong
  5. Sep 14, 2009 #4


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    For jumping diver t = 0.92 s to reach the watr.
    For walking diver t = 0.74.
    So who reaches the water first?
    Using the first equation you can find their final velocities.
  6. Sep 16, 2009 #5
    Based on the shorter time for the walking diver, he reaches the water first.

    I got the final velocity of the step off diver correct, but the final velocity of the jumping diver was wrong, this is what I did.

    Vf^2=Vi^2 + 2ad
    Vf^2= 0 + 2(9.8m/s^2)(2.70m+1.6m)
    Vf= 9.18m/s

    I think my d in the equation is wrong, what do you think?
  7. Sep 16, 2009 #6


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    Initial velocity of the jumper is 1.6 m/s in the upward direction. When he crosses the starting point, his velocity is 1.6 m/s.
    So vf = vi + g*t where t = 0.74 s
  8. Sep 16, 2009 #7
    Alright so, Vf= Vi +gt
    Vf= 1.6m/s + (9.8m/s^2)(0.74s)(0.74s)
    Vf= 1.6 + 7.252
    Vf= 8.852m/s

    But if I'm dealing with the jumping diver, why am I using the time from the diver that's walking off? shouldn't I use t=0.92s?
  9. Sep 16, 2009 #8
    That was a typo on my part, I didn't mean to type it twice. But yeah, why am I using the t=0.74s. Why am I not using t= 0.92s?
  10. Sep 16, 2009 #9
    wait Im confused, how is it the same, I previously found 2 different times from which both swimmers will hit the water.
  11. Sep 16, 2009 #10


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    For walker
    vf = 0 + g*t = 9.8* 0.7274
    For jumper
    vf = vi - g*t Here vi = 1.6 and t = 0.9233
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