Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Confusing Kinematics and Acceleration

  1. Oct 13, 2003 #1
    Hello everyone. I've been at this for an hour now, so I think it's about time I asked for help.

    We have a person on the surface of a planet with an unknown gravitational acceleration. The only measuring instrument available is a very accurate watch.

    The person is on top of a cliff, and drops a rock from the top of it. It takes 4.15s for the rock to fall from the top of the cliff to the ground below. It travels X meters in total.

    Afterwards, the person takes the rock, and throws it so that it travels up 2 meters above the top of the cliff, then it falls down to the ground. It travels (4 + X) meters in total Here is a simple depiction:
    Case 1)
    ||
    ||
    ||
    \/

    Case 2:
    //\\
    || ||
    || ||
    || ||
    ||
    ||
    ||
    \/

    This shouldn't be that hard, but I'm still not getting it. Any help would be greatly appreciated. Thank you.
     
  2. jcsd
  3. Oct 13, 2003 #2
    OK
    In the first case, when the person drops the rock, the distance from the cliff to the bottom is:
    d=a*(((4.15)^2)/2)
    In the second case, when the person launchs the rock upwards, we can put:
    2+d=(a*(t^2))/2------>d=a*(t^2)/2-2
    Equalling the 2 d, you eliminate a and get t, that is the time that the rock lapses in the second launch from the upper part of its trajectory to reach the ground
    The solution of the rest of the problem is straightforward
     
  4. Oct 13, 2003 #3
    Sorry, but i really don't understand your response. I understand that for the first part, where the rock drops, we can use 0.5(a)(t^2), but for the second part, you used the same equation displayed in a different form (with +2m on one side and -2m on the other). I'm really at a loss on this question.
     
  5. Oct 13, 2003 #4

    Integral

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The general equation you need is

    s = .5at2 + v0t+x0

    Where v0 is the initial velocity and
    x0 is the initial position.
    s is the position of the rock at t seconds.

    Now we need to identify the parameters. For the first part the rock is dropped so its initial velocity is 0 (v0=0) and the initial postion = X (x0=X)
    we need the time when s=0
    So set up the first equation using these parameters.

    0= .5at2+X

    so solve for a

    a = -2X/t2

    Now the second part cam be set up several different ways.
    Since the rock is thrown up it is given some initial velocity v0, which we are not given, we only have the distance traveled.

    we have.
    2=.5aT2+ v0T
    where T would be the time required to reach the top of motion of the rock, we do not have T, nor v0 so this does not help us.

    We could write
    0 = .5aΤ2+v0Τ + X

    Where Τ is the time required to reach the ground after being thrown up. Again we do not have Τ or v0. Once again we are stuck.
    Or We could write

    0= .5aτ2+X+2,
    this equation results from the fact that the rock has zero velocity 2m above the edge of the cliff and falls to the ground. Here we would need τ, the time for the rock to fall from its peak of travel to the foot of the cliff, once again we do not have this information.

    We have only information on how far the rock travels, we do not have an initial velocity, or a time of travel. Without these bits of information the problem cannot be solved.
    Did you simply forget to include a time for part 2 or was it not given?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Confusing Kinematics and Acceleration
  1. - Kinematics - (Replies: 1)

Loading...