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Math Help - a rock is thrown straight upward with an initial velocity of 50 m/s from a point 100

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    a rock is thrown straight upward with an initial velocity of 50 m/s from a point 100

    a rock is thrown straight upward with an initial velocity of 50 m/s from a point 100 m above the ground. If the acceleration of the rock at any time t is a= -10 m/ s^2 , what is the maximum height of the rock (in meters)?

    here's my attempt:
    a(t)= -10t

    take anti-derivative

    v(t)= -5t^2+c

    v(0)= -5(0)^2+c
    50= c

    v(t)= -5t^2+50

    s(t)= \frac{-5}{3}t^3+50t+100

    then i dont know what to do from here
    Last edited by yoman360; March 27th 2010 at 11:06 AM.
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  2. #2
    Member mathemagister's Avatar
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    Quote Originally Posted by yoman360 View Post
    a rock is thrown straight upward with an initial velocity of 50 m/s from a point 100 m above the ground. If the acceleration of the rock at any time t is a= -10 m/ s^2 , what is the maximum height of the rock (in meters)?
    a(t) = -10

    \int a(t) dt = v(t) + C = -10t + C

    Initial velocity is v(0), which is 50

    v(0) = -10 \cdot 0 + C \implies v(0) = C = 50

    v(t) = -10t + 50

    The maximum height of the rock occurs when the velocity is 0. (Think about it, that is the time when the rock is neither moving up nor down)

    v(t) = -10t + 50 = 0 at max height

    t = 5

    Now you know when the maximum occurs, at t = 5

    \int v(t) dt = s(t) + C = -5t^2 + 50t + C

    The initial position is 100. So s(0) = C = 100

     s(t) = -5t^2 + 50t + 100

    Maximum = s(5)

    s(5) = -125 + 250 + 100 = 225

    So the answer is 225m.

    Hope that was helpful
    Last edited by mathemagister; March 27th 2010 at 11:11 AM. Reason: mathtype typo
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