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Homework Help: Motion in a straight line HELP

  1. Jun 9, 2008 #1
    Motion in a straight line HELP!!!!

    Here's my problem...

    A rock is dropped (from rest) from the top of a 87.5-m-tall building. How far above the ground is the rock 1.2 s before it reaches the ground?

    .....so far this is what I came up with


    t=sqrt(2y/g) = sqrt((2*87.5)/9.8) = 4.23 s <-- I believe this is the time for the entire drop?

    Then I subtracted 4.23-1.2 = 3.03 s <---- so thats the time the rock falls before the point at which i need to calculate the height.

    After that i get lost and can't seem to come up with a sensible answer.

    If someone could walk me through this I'd really appreciate it!
  2. jcsd
  3. Jun 9, 2008 #2


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    Homework Helper

    well you are going correct with the formula


    since it starts at rest,u=0. and the acceleration a=g


    so that is correct.

    you want to find s,when t=1.2. Just put it into the formula and you'll get s when t=1.2
  4. Jun 9, 2008 #3
    OK so if I plug 1.2 s into the equation I get this....

    = 7.056 m/s


    Now I still don't get how I figure out how high off the ground the rock is at 1.2 seconds?
  5. Jun 9, 2008 #4


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    Homework Helper

    s is displacement. so the unit is m.

    So 7.056 is the height from the top of the building to the point when t=1.2seconds.

    The height of the entire building is 87.5m.

    So the height from the ground would just be (height of building)-(distance when t=1.2)
  6. Jun 9, 2008 #5
    Hey goaliejoe,
    Your on the right track but use your time that you found (3.03) and plug that into the formula
    Y = Yo - (1/2)at^2. Your y naught is the starting height.
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