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Free fall

  1. Sep 15, 2009 #1
    1. The problem statement, all variables and given/known data
    A pen falls from the top of a roof. We know that it takes 1.5 seconds to fall the last 1/4 distance. Find the height of the roof from the ground.


    2. Relevant equations
    x=x_0 + v_0 * t + 1/2 gt^2
    v=v_0 + gt



    3. The attempt at a solution
    Since the problem does not say that the pen falls from rest, I don't know what v_0 should be. Also, how can I make use of the time that it travels the last 1/4 distance.
    Any help please.
     
  2. jcsd
  3. Sep 15, 2009 #2

    rl.bhat

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    A pen falls from the top of a roof. So its initial velocity = 0. Let t1 be the time to reach the ground ( h ) and t2 be the time to cover the distance 3h/4. Write down two equations;
    h = 1/2*g*t1^2....(1)
    3h/4 = 1/2*g*t2^2 ......(2)
    You can rewrite the two equation in terms of t1 and t2.
    Since t1 - t2 is given, you can find h.
     
  4. Sep 16, 2009 #3
    hello rl.bhat
    (1)-(2) gives h/4=1/2*g(t1-t2)*(t1+t2) and so t1+t2 can be easily calculated but what from there?
    greetings Janm
     
  5. Sep 16, 2009 #4

    rl.bhat

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    Divide (1) by (2) you get
    4/3 = t1^2/t2^2
    2/(3)^1/2 = t1/t2
    (2/1.732) - 1 = (t1/t2) - 1
    Solve this equation to find t2.
     
  6. Sep 17, 2009 #5
    Hello rl.bhat
    For sqrt(3) I tend to use 26/15. So your third equation becomes:
    30/26=15/13=t1/t2
    I am getting a hunch why you write the fourth equation, which becomes:
    2/13=(t1-t2)/t2
    is that correct?
    greetings Janm
     
  7. Sep 17, 2009 #6

    rl.bhat

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    Yes. It is correct. Now put the value of (t1 - t2) and find the value of t2.
     
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