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Kinematics Question

  1. Sep 11, 2005 #1
    Here is the question I am trying to awnser:

    A ball is thrown straight up and reaches a maximum height in 6.81 s.

    What was the ball's initial speed?

    Keeping in mind that we are not yet dealing with any sort of friction...

    So for problems like this I do the following:

    Given:
    Xo - 0
    X - ? (this might be zero I'm not sure)
    Vo - ?
    V - ? (this might be zero I'm not sure)
    a - 9.8 m/s
    t - 6.81 at it's maximum height

    I need to solve for Vo. I'm not sure how to do this though.

    I know I should use one of the equations of:

    [​IMG]

    But I have no idea beyond that...

    I tried the V=Vo + at with V as zero and t as 13.62 seconds (I doubled it beacuse I thought I should in order to get the complete time) and when I did that I got Vo as a negative...

    And I know it's not a negative...

    Any help???
     
  2. jcsd
  3. Sep 11, 2005 #2

    mrjeffy321

    User Avatar
    Science Advisor

    You were on the right track with the formula you used.

    V = Vi + at
    Velocity = The initial Velocity + (acceleration * time)

    The point when the object reaches its maximum height is when it has zero upward velocity (V = 0). While the object is traveling upward, it is undergoing a constant downward acceleration due to gravity (a = -9.81).
    and the time it took to reach it max height is given (t = 6.81)

    so,
    0 = Vi + (-9.81 meters/second^2)(6.81 seconds),
    now solve for the initial velocity.
     
  4. Sep 11, 2005 #3
    There is no need to double the time, as V=0 after 6.81 seconds. (or if you want to make life more difficult put V=-Vi and t=13.62)
     
  5. Sep 11, 2005 #4
    So it's initial speed is 66.738 m/s right???
     
  6. Sep 11, 2005 #5
    I'm pretty sure that's correct (I got the same answer), except in fewer sig figs of course :)
     
  7. Sep 11, 2005 #6
    Ok I'm pretty sure this is the last question for now...

    It reads as this:

    A ball is dropped from rest at point O (height unknown). After falling for some time, it passes by a window of height 3.4m and it does so during tAB = .45 s.

    The ball accelerates all the way down; let Va be its speed as it passes the window's top A and Vb its speed as it passes the window's bottom B.

    How much did the ball speed up as it passed the window; i.e., calculate Triange(that's the delta sign right) Vdown = Vb - Va? Awnser in units of m/s.

    Then, calculate the speed Va at which the ball passes the window's top. Awnser in units of m/s.

    It has a diagram that looks like this: http://img11.imageshack.us/img11/3634/ff5ge.png


    I honestly have no clue where to start on this...

    I'm not even sure what it is asking me to find out...

    I think I may have gotten the awnser for the first question, but I doubt it...

    I did, V=Vo + at which is, with the subsitution:

    V = 0 + 9.8 m/s(.45 sec)

    Which comes out to be V = 4.41 m/s. However, I'm not sure what that V is. Is that the Velocity the ball has at the top of the window (in the diagram A)? Or is it something else? Did that solve the second question not the first?

    Help would be great, thanks in advance!!!
     
  8. Sep 11, 2005 #7
    Ohp, nevermind...

    Got it...
     
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