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Physics Homework Problem

  1. Feb 11, 2005 #1
    A diver springs upward with an initial speed of 1.7 m/s from a 3.0 m board.

    a. Find the velocity with which he strikes the water. (Hint: When the diver reaches the water, his displacement is y = -3.0 m (measured from the board), assuming that the downward direction is chosen as the negative direction.)

    b. What is the highest point he reaches above the water?


    I tried using this equation, but my answer was wrong. Can someone tell me what I did wrong?

    Part a:
    d=v*t + 1/2 at^2
    -3.0=1.7t + 1/2(-9.80)t^2
    4.9t^2 - 1.7t -3.0=0
    t= 0.98s

    v=d/t
    v=3/0.98
    v=3.06m/s

    I didn't do Part b because Part a was wrong!!
     
  2. jcsd
  3. Feb 11, 2005 #2
    When I did this question, I first noticed to use energy. For me, that's more comfortable.
    However, your method also is a good method, even better than mine.
    t is correctly found.

    "v=d/t
    v=3/0.98
    v=3.06m/s" is wrong.
    Try to find out what's the exactly meaning of the v in v=d/t.
    You are going to find a particular v but not that one.
     
  4. Feb 11, 2005 #3

    dextercioby

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    For the second question:HINT:Use Galiei's formula,the one with the squares of velocities.

    Daniel.
     
  5. Feb 12, 2005 #4

    xanthym

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    A) Hint: Remember the initial velocity v(0):

    [tex]: \ \ \ \ \ v(t) = v(0) + at [/tex]

    B) Formula with "velocities squared" is equivalent to Energy approach for constant acceleration "a":

    [tex]: \ \ \ \ \ v^2(t) - v^2(0) = 2a(d(t) - d(0)) [/tex]

    [tex]: \ \ \ \ \ \frac {mv^2(t)} {2} - \frac {mv^2(0)} {2} = (ma)(d(t) - d(0)) [/tex]

    [tex] :: \ \ \ \ \ \Delta K.E. = (Force)(\Delta Distance) [/tex]


    ~~
     
  6. Feb 14, 2005 #5

    dextercioby

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    Double posting is not allowed...:grumpy:

    Daniel.
     
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