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Thread: Problem Involving Motion

  1. #1
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    Problem Involving Motion

    A canister is dropped from a helicopter hovering 500m above the ground. Unfortuantly its parachute does not deploy. It has been designed to withstand an impact velocity of 100m/s. Will it burst or not?

    Not sure what to do.
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  2. #2
    Senior Member Spec's Avatar
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    Check if $\displaystyle mgh\leq\frac{mv^2}{2}$ where $\displaystyle mgh$ is the potential energy at the drop point and with the ground as zero. Assuming my understanding of impact velocity in this question is correct.
    Last edited by Spec; May 19th 2009 at 07:30 PM.
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  3. #3
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    No idea how to use that
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  4. #4
    Senior Member Spec's Avatar
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    Just plug in the values of the variables into the inequality and see if it holds.

    $\displaystyle h=500$ (height)
    $\displaystyle v=100$ (velocity)
    $\displaystyle g=9.8$ (standard gravity)

    $\displaystyle m > 0$ so you can just divide it out.

    $\displaystyle 9.8 \cdot 500 \leq 10 \cdot 500$ so it will not burst.

    This is not exactly an answer to a calculus question, so I don't think I have interpreted the impact velocity correctly.
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  5. #5
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    Quote Originally Posted by Dickson View Post
    A canister is dropped from a helicopter hovering 500m above the ground. Unfortuantly its parachute does not deploy. It has been designed to withstand an impact velocity of 100m/s. Will it burst or not?

    Not sure what to do.
    For simple equation of motion without air resistance, use:

    v^2 = U^2 + 2*a*s

    v^2 = 0 + (2*9.8*500)

    v^2 = 9800

    v = sqrt (9800) = 98.99 m/s

    Pretty close then!

    If you factor in air resistance, the velocity at impact will be slower
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