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Homework Help: Professor leaps off a helicopter, ball thrown in the air Kinematics

  1. Sep 15, 2010 #1
    Hey all, I've been lurking for a while to try and see if these problems were solved already but I couldn't find anything so here I am posting!

    Question #2:A ball is thrown straight upward from the ground. The ball passes a window 4.35 m above the ground and is seen to descend past that same window 3.68 s after it went by on the way up.

    1. What is the total length of time the ball is in the air?


    2. Relevant equations
    [tex]
    v = v_0 + a t
    [/tex]
    [tex]
    x = x_0 + v_0 t + (1/2) a t^2
    [/tex]
    [tex]
    v^2 = v_0^2 + 2 a \Delta x
    [/tex]
    and Quadratic formula

    3. The attempt at a solution
    I tried finding the velocity, max height and then plugging it in the quadratic formula but it is not the right answer. I'm not sure how else to do it? Any help would be greatly appreciated
     
    Last edited: Sep 16, 2010
  2. jcsd
  3. Sep 15, 2010 #2

    cepheid

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    Gold Member

    Welcome to PF,

    Why are the values in red negative?

    For the last part, the trip can be divided into three stages:

    - free fall
    - deceleration
    - constant velocity

    All you need to do is calculate the duration of each stage. For the final stage, the amount of time it takes her to descend at a constant velocity just depends upon the her height above the ground at the beginning of that stage.
     
  4. Sep 16, 2010 #3
    Ok got it, thanks! But can someone help me out with the second problem too?
     
  5. Sep 16, 2010 #4

    cepheid

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    Staff Emeritus
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    Gold Member

    How about this? Call the time at which the ball reaches a height of 4.35 m time t1. Substituting this into the middle equation in your relevant equations, you only have one other unknown, which is v0. Therefore, you can solve for v0 in terms of t1.

    Now, you know from the inherent symmetry of the situation that the time it takes to go up from 4.35 m to the max height should be equal to the time that it takes to come back down to 4.35 m from the max height -- it's (3.68/2) = 1.84 seconds in either direction. Therefore, you can substitute your expression for v0 in terms of t1 into the first equation in your list of relevant equations. Here, you set v = 0 and t = t1 + 1.84 s. This equation then gives you t1.
     
  6. Sep 16, 2010 #5
    Wow thank you very much! I finally figured them both out! Thank you once again!!
     
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