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Rock tossed straight up-Velocity question

  1. Feb 8, 2009 #1
    1. The problem statement, all variables and given/known data

    A rock is tossed straight up with a speed of 21 m/s. When it returns, it falls into a hole 15 m deep. What is the rock's velocity as it hits the bottom of the hole? How long is the rock in the air, from the instant it is released until it hits the bottom of the hole?

    2. Relevant equations

    d = vit + (0.5)at^2

    3. The attempt at a solution

    d= 15 m
    a= 9.8
    vf=
    vi= -21 m/s
    t=

    I was able to find the time by plugging in:

    15= (-21)t+4.9t^2

    which was a quadratic, so it turned out to be 4.9 seconds

    I'm just not sure about how to find the velocity at the bottom of the hole. I've been trying for about an hour now and this one problem is driving me mad.

    I think the initial velocity would change to 0 and the acceleration due to gravity would still be the same. I tried using v=vo+at but I got a wrong answer. (V=0+9.8*4.9=48)

    I'm using masteringphysics so I can try answers to see if they are correct, but I'm on my last attempt and I'm pretty desperate.

    Please explain to me what I am doing wrong and help me solve this problem!
     
  2. jcsd
  3. Feb 8, 2009 #2

    LowlyPion

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    Homework Helper

    Welcome to PF.

    You almost had it.

    It is V = Vo - g*t

    According to the statement though the initial Vo is 21 m/s UP.

    So your answer will likely need to be expressed as minus, as the equation yields.

    V = 21 - 9.8*(4.9)
     
  4. Feb 8, 2009 #3

    Redbelly98

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    Staff Emeritus
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  5. Feb 8, 2009 #4
    You only want to consider the downward motion of the rock starting from it's highest point (when it stops the upward motion). The only acceleration will be due to gravity. Don't forget to add the extra depth for the hole.
     
  6. Feb 8, 2009 #5
    Thank you so much! I really appreciate it!
     
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