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Motion of a body under acceleration of gravity

  1. May 23, 2007 #1
    I have been looking at this question from a Mechanics Chapter:-

    A stone is thrown vertically upwards with a velocity of u metres per second. It passes a ledge in t1 seconds and repasses it t2 seconds after the start. Find the height of the ledge?

    Take g = 9.8 metres per second sq

    This is the question in entirety.

    I have tried using

    s = ut + 1/2 a t^2 substituting g in as

    s = ut - 1/2 g t^2

    If the height is the same at both these times then I have assumed that:-

    ut1 -1/2 g t1^2 = ut2 - 1/2 g t2^2

    Then factoring out t1 and t2

    t1(u - 1/2 g t1) = t2(u - 1/2 g t2)

    then failed to see how this could find s!

    Then tried rearranging the original formula in terms of u and t1, so:

    u = (s + 1/2 g t1^2)/t1

    and substituting into

    s = ut2 - 1/2 g t2^2

    but couldn't proceed.

    Can somebody help? I am assuming that I need to find s in terms of g, t1 and t2 rather than an actual value.
     
  2. jcsd
  3. May 23, 2007 #2

    Dick

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    That's REALLY the whole question? Its a bit mysterious what they are really after. Do you really want to just eliminate u from the expression for s? That seems to be what you are trying to do.
     
  4. May 23, 2007 #3
    Why can't you just use the formula
    d = Vo * t + 1/2 a * t^2
    where
    d = height
    Vo = initial speed = u
    t = time = t1
    a = 9.8
    t = t1
    I don't think you can find out the actual value. This would be how you would calculate d using u and t1. Wouldn't this be enough (as I am assuming I actually know the value of the speed and of the first time - they are given in the problem)?
     
  5. May 23, 2007 #4

    Dick

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    That's why I'm groping for what the problem is actually asking.
     
  6. May 24, 2007 #5
    That honestly is the whole question word for word. Unfortunatly I dont even have the answer to work backwards from either.

    Re: Dick - Yes that was what I was trying do - to simplify S in terms of g and t1 or t2. Then I was wondering if I could even eliminate u, as I initially assumed that the initial velocity for both equations (s for t1 and s for t2) is the same but thinking about it, this might not be true.

    Or maybe find t1 in terms of t2 (or vice versa) and then substitute into...but not sure what!

    Re: husky88 - I agree this could potentially be the answer, just that I thought that in a question like this, details like the time t2 arent provided unless they are intended to be used in the final answer.
     
  7. May 24, 2007 #6

    Dick

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    You can eliminate any one of u,t1,t2. One of them is redundant. E.g. What's the velocity at time (t1+t2)/2? So u/g=(t1+t2)/2. That's all three in one simple equation. Know any two and you know the third. I'm tempted to just write down s=u*t1-g*t1^2/2 and call it a day.
     
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