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Homework Help: Phy 111

  1. Sep 10, 2008 #1
    1. The problem statement, all variables and given/known data
    A stone is launched straight up by a slingshot. Its initial speed is 20.2 m/s and the stone is 1.40 m above the ground when launched. Assume g = 9.80 m/s2.

    A:How high above the ground does the stone rise?
    B:How much time elapses before the stone hits the ground?

    2. Relevant equations
    Y=Vit+ 1/2ayt2

    3. The attempt at a solution
    I have tried the first formula with the time from the second and can't get the answer. I have tried everything I can think of and was even working with others at school to try and get this one. I have seen the other problem identical to this listed but I couldnt understand it. Please Help
    Last edited: Sep 10, 2008
  2. jcsd
  3. Sep 10, 2008 #2


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    Homework Helper

    Welcome to PF.

    I'm sure you meant this equation to be:

    Maybe that will be all you need?

    ay = -g
    Y + 1.4 = Ymax
    Last edited: Sep 10, 2008
  4. Sep 10, 2008 #3
    Thanks for the welcome. And yes I had that, I just input the equation wrong into the computer.

    When I go through and try an answer I have gotten 2.06 for the time it takes for it to stop rising which I would think i could just input into the formula to have it be some thing like y=1.40m+20.2m/s(2.06s)+1/2(-9.8m/s2)(2.06s)2. I end up getting 22 but I dont think that is the answer, it just doesnt make sense with it rising for 2 seconds with an initial velocity of 20m/s...
  5. Sep 10, 2008 #4


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    Homework Helper

    Use the equation I suggested first.

    That's (02 - (20.3)2)/2(-9.8) = Y
    where your final velocity is 0.
    Then to that number you add the additional 1.4 m.

    Now use the height Y to figure Time to MAX and then use Y + 1.4 to figure time to fall.. These two times you can calculate simply with Y = 1/2 g t2

    Then add the two time results together. That's Total time.
  6. Sep 10, 2008 #5
    Thank you! Finally got it.
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