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Thread: stone dropped

  1. #1
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    stone dropped

    a stone is dropped from rest from the top of a tower. In the last second of its motion it falls through a distance 1/4 of the height of the tower. find the height of the tower.

    I am trying to solve this using UVAST equations but so far I can not do it.
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  2. #2
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    Re: stone dropped

    Thanks from topsquark
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  3. #3
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    Re: stone dropped

    The acceleration due to gravity is -9.8 m/s^2. Starting from rest, its speed after t seconds is -9.8t m/s. If it started from height h, its height after t seconds is -4.9t^2+ h m. It will hit the ground when -4.9t^2+ h= 0 or t= sqrt(h/4.9) seconds.

    A second before that is t= sqrt(h/4.9)- 1 and at that time it will have gone -4.9(h/4.9- 2sqrt(h/4.9)+ 1)+ h. So in the last second it went a distance h- 4.9(h/4.9- 2sqrt(h/4.9)+ 1) and that is equal to (1/4)h. You need to solve h- 4.9(h/4.9- 2sqrt(h/4.9)+ 1)= h/4.
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  4. #4
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    Re: stone dropped

    where do you get -4.9(h/4.9- 2sqrt(h/4.9)+ 1)+ h from?

    I let u=o and s=3h/4 and a=g and t= sqrt(h/4.9)- 1

    using s=ut+.5t^2 I get some unsolvable equation.
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  5. #5
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    Re: stone dropped

    Quote Originally Posted by edwardkiely View Post
    where do you get -4.9(h/4.9- 2sqrt(h/4.9)+ 1)+ h from?

    I let u=o and s=3h/4 and a=g and t= sqrt(h/4.9)- 1

    using s=ut+.5t^2 I get some unsolvable equation.
    The initial height is $h$. The starting velocity is $v$. The starting acceleration is $a$. Then:

    $s = h+vt+\dfrac{a}{2}t^2$

    As you pointed out, $v=0, a = -g$ (note the negative sign for the acceleration indicating the direction of movement is down). You are trying to solve for $h$.

    Plugging in the known variables, you have:

    $s = h-4.9t^2$

    At time $t, s=0$. So, plugging in, you have:

    $0 = h-4.9t^2$
    $4.9t^2 = h$
    $t = \sqrt{\dfrac{h}{4.9}}$

    At time $\sqrt{\dfrac{h}{4.9}}-1$, you have $s = \dfrac{h}{4}$. Plugging in:

    $\dfrac{h}{4} = h-4.9\left(\sqrt{\dfrac{h}{4.9}}-1\right)^2$

    Subtract $\dfrac{h}{4}$ from both sides, add $4.9\left(\sqrt{\dfrac{h}{4.9}}-1\right)^2$ to both sides:

    $\dfrac{3h}{4} = 4.9\left(\dfrac{h}{4.9}-2\sqrt{\dfrac{h}{4.9}}+1\right) = h-2\sqrt{4.9h}+4.9$ (this is what HallsofIvy showed you)

    Subtract $\dfrac{3h}{4}$ from both sides:

    $0 = \dfrac{h}{4}-2\sqrt{4.9h}+4.9$

    Now, we have a quadratic of the variable $\sqrt{h}$.

    $\sqrt{h} = \dfrac{2\sqrt{4.9} \pm \sqrt{ (2\sqrt{4.9})^2 - 4\left( \dfrac{1}{4} \right) (4.9) } }{2\left( \dfrac{1}{4}\right) }$

    $\sqrt{h} = 4\sqrt{4.9} \pm 2\sqrt{3}\sqrt{4.9}$

    $\sqrt{h} = (4+2\sqrt{3})\sqrt{4.9}$ or $\sqrt{h} = (4-2\sqrt{3})\sqrt{4.9}$

    Square both sides to get:

    $h = 4.9(4\pm 2\sqrt{3})^2$

    This will give you two possible solutions for the initial height of the tower. Both will be correct (since $4-2\sqrt{3}>0$).
    Thanks from topsquark and edwardkiely
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