1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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


    User Avatar
    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Motion in a straight line HELP
  1. Straight line motion (Replies: 4)