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Math Help - Motion with Uniform Acceleration

  1. #1
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    Motion with Uniform Acceleration

    Though id post this one here becuase its partly mechanics so physcis based etc. This thing is im working with SuVat as in S=displacement u=uniform acceleration V=velocity a=acceleration and t=time. Ive manged to answer most of the questions I have on this work but have become stuck on the final one and was wondernig if anyone can help. The question is A stone fall past a window, 3m high, in 0.5s. Taking acceleration due to gravity, g (a) = 9.8ms-2, find the height from which the stone fell. It would be great help if you could help me out. The formulas I need to use are v=u+at s=ut+1/2at squared. v squared = u squared + 2as and s=1/2(u+v)t. sorry if it not clear.

    Thankyouu!!
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  2. #2
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    Quote Originally Posted by NathanBUK View Post
    Though id post this one here becuase its partly mechanics so physcis based etc. This thing is im working with SuVat as in S=displacement u=uniform acceleration V=velocity a=acceleration and t=time. Ive manged to answer most of the questions I have on this work but have become stuck on the final one and was wondernig if anyone can help. The question is A stone fall past a window, 3m high, in 0.5s. Taking acceleration due to gravity, g (a) = 9.8ms-2, find the height from which the stone fell. It would be great help if you could help me out. The formulas I need to use are v=u+at s=ut+1/2at squared. v squared = u squared + 2as and s=1/2(u+v)t. sorry if it not clear.

    Thankyouu!!
    You have to combine 2 equations:

    \left|\begin{array}{rcl}s+3&=&\frac12 \cdot 9.81 \cdot (t+0.5)^2 \\ s&=& \frac12 \cdot 9.81 \cdot t^2 \end{array} \right.


    Expand the first equation and subtract columnwise the second equation from the first one. You'll get an equation in t. Solve for t.

    I've got t \approx 0.3616\ s which means that the stone starts to fall 0.64 m above the window.
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  3. #3
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    Quote Originally Posted by NathanBUK View Post
    Though id post this one here becuase its partly mechanics so physcis based etc. This thing is im working with SuVat as in S=displacement u=uniform acceleration V=velocity a=acceleration and t=time. Ive manged to answer most of the questions I have on this work but have become stuck on the final one and was wondernig if anyone can help. The question is A stone fall past a window, 3m high, in 0.5s. Taking acceleration due to gravity, g (a) = 9.8ms-2, find the height from which the stone fell. It would be great help if you could help me out. The formulas I need to use are v=u+at s=ut+1/2at squared. v squared = u squared + 2as and s=1/2(u+v)t. sorry if it not clear.

    Thankyouu!!
    I'm assuming the stone was released from rest.

    the d = distance from where the stone was dropped to the top of the window

    t = elapsed time from when the stone was released until it gets to the top of the window

    d = \frac{1}{2}gt^2

    d+3 = \frac{1}{2}g(t+0.5)^2<br />

    substitute \frac{1}{2}gt^2 for d in the second equation ...

    \frac{1}{2}gt^2 + 3 = \frac{1}{2}g(t+ 0.5)^2

    \frac{1}{2}gt^2 + 3 = \frac{1}{2}g(t^2 + t + 0.25)

    \frac{1}{2}gt^2 + 3 = \frac{1}{2}gt^2 + \frac{1}{2}gt + \frac{1}{8}g

    3 =  \frac{1}{2}gt + \frac{1}{8}g

    24 = 4gt + g

    t = \frac{24-g}{4g}

    d = \frac{1}{2}g\left(\frac{24-g}{4g}\right)^2
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  4. #4
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    Oh thankyouu soo much. One things for sure I deff havent picked this up well in class that working out loooks sooo confusing but Im sure ill get the correct answer if I follow youre working out.
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