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(Mechanics) Getting a distance in a Ball Drop question!

  1. Sep 13, 2011 #1
    1. The problem statement, all variables and given/known data

    A ball is dropped from the roof of a building. An observer looking outside the window sees the ball just outside. If the window is 3.99 m tall and the ball is in view for 0.897 s, how far below the roof is the bottom of the window?



    2. Relevant equations

    "Kinematics Equations
    average velocity:
    vave=Δx/Δt


    average acceleration:
    aave=Δv/Δt


    Uniform Acceleration:

    vave=(vi+vf)/2


    Three key variables (displacement, time, velocity) lead to three key relationships relating each pair:

    velocity & time:
    v=v0+at


    displacement & time:
    x=x0+v0t+(1/2)at2


    velocity & displacement:
    v2=v20+2aΔx"
    source : Physicsforums.com



    3. The attempt at a solution

    I do not know how to even start this question from..
    For example, what equations to use or what numbers do I need to get in order to solve this question.
    Please help me with this one guys, thank you very much!
     
  2. jcsd
  3. Sep 13, 2011 #2
    If I get stuck in any physics problem, the first thing I do is make a list of knowns and unknowns. What are your knowns?
     
  4. Sep 13, 2011 #3
    mm.. knowns are..
    I actually am confused with the "window is 3.99m tall".. if it means the window itself is 3.99 m tall OR it's up at the height of 3.99m from the ground.. if it means itself is 3.99m tall,
    I think knowns would be the
    h of window = 3.99
    t of ball while in sight of the person = 0.897
    V0 = 0 m/s
    a = -9.80 m/s^2

    ??
     
  5. Sep 13, 2011 #4
    bump.. please help :/
     
  6. Sep 13, 2011 #5
    There are a couple of "keys" to this problem which will help you cancel some items in the equations. If you draw a picture of this problem you will see that the total distance traveled by the ball will be y + 3.99m and also the total time will be t + 0.897s.

    Now for the ball to travel from the top of the window to the bottom of it, it needed a certain starting velocity and then an ending velocity (because gravity is adding to the velocity). The starting velocity at the top of the window is actually the same velocity as the total distance traveled MINUS 3.99m. Then that makes the velocity at the bottom of the window the same as the total distance traveled.

    I'm not going to do the math but that's the conceptual part that should get you started.
     
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