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Velocity of 1-dimensional projectile

  1. Jan 25, 2009 #1
    1. The problem statement, all variables and given/known data
    A projectile is launched straight upward at +63.0 m/s from a height of 85.0 m, at the edge of a sheer cliff. The projectile falls, just missing the cliff and hitting the ground below.

    Maximum height = 202.5m

    1) Find the time it takes to hit the ground at the base of the cliff.
    2) Find its velocity at impact

    2. Relevant equations
    xf=xi+vit+1/2at2
    vf=vi+at


    3. The attempt at a solution
    For #1 I tried this:
    xi=202.5 xf=0 a=-9.80 vi=0
    0=202.5+0+1/2(-9.8)t2
    -202.5=-4.9t2
    41.327=t2
    t=6.429

    But that's not the right answer

    For #2 I tried this: (It depends on the answer to #1)
    Vf=0+(-9.8)(6.429)
    Vf=-63.004

    But that's wrong too since #1 is wrong. What did I do wrong in #1?
     
  2. jcsd
  3. Jan 26, 2009 #2

    Gib Z

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    Homework Helper

    By putting x_i = 202.5, you are only solving for the time it takes to fall from the top of its motion, down to the ground. You have to account for the time it took to get up there!
     
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