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Math Help - Kinematics

  1. #1
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    Kinematics

    Question:

    A stone is thrown vertically downwards from the top of a cliff. Its height h metres above the beach below at time t seconds is given by.
    h= (10- sqrt 5t) (10 + sqrt 5t)

    show that acceleration is constant and state its value.

    hence find the time when it hits the beach.

    The answer is -10 ms-2 however I don't know how to manipulate the above expression.

    sqrt 5t should be equal which would give a value of -25 and not 10, which brings me to wonder if the question is supposed to say sqrt 5 times t, would it make a difference? 5t^0.5 however when one multiplies the two and differentiates again would give -25.
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  2. #2
    Super Member redsoxfan325's Avatar
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    Well, (10-\sqrt{5t})(10+\sqrt{5t}) = 100-5t. So the stone hits the ground at t=20.

    ---------------

    However, the problem probably means h=(10-t\sqrt{5})(10+t\sqrt{5}) = 100-5t^2, so the stone hits the ground at t=\sqrt{20}=2\sqrt{5}.

    Thus, velocity = \frac{dh}{dt}=-10t~m/s and acceleration = \frac{d^2h}{dt^2} = -10~m/s^2
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  3. #3
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    Quote Originally Posted by Mouseman View Post
    Question:

    A stone is thrown vertically downwards from the top of a cliff. Its height h metres above the beach below at time t seconds is given by.
    h= (10- sqrt 5t) (10 + sqrt 5t)

    show that acceleration is constant and state its value.

    hence find the time when it hits the beach.

    The answer is -10 ms-2 however I don't know how to manipulate the above expression.

    sqrt 5t should be equal which would give a value of -25 and not 10, which brings me to wonder if the question is supposed to say sqrt 5 times t, would it make a difference? 5t^0.5 however when one multiplies the two and differentiates again would give -25.
    h = (10 - \sqrt{5} t)(10 + \sqrt{5} t)

    h = 0 at t = \frac{10}{\sqrt{5}} sec

    expand the expression for h ...

    h = 100 - 5t^2

    \frac{dh}{dt} = v = -10t

    \frac{dv}{dt} = a = -10
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