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Thread: Derivatives, Projected object

  1. #1
    Senior Member vaironxxrd's Avatar
    Nov 2010

    Derivatives, Projected object

    As you guys might know I don't want an answer, I would just like some guidance.

    An object is projected upward from the ground with an initial velocity of 128 feet per second. The height s at the end of t seconds is s = $\displaystyle 128t - 16t^2$.

    When does it reach its maximum height and what is it?

    The object reaches maximum height when the velocity is 0.

    $\displaystyle v = D_x(128t-16t^2)$ = $\displaystyle v = 128-32t$
    t = 4
    s = 128(4)-32(4) = 256 ft

    When does it hit the ground and with what velocity?

    The object will hit the ground when the height is 0

    Therefore $\displaystyle 128t -16t^2 = 0$

    I got stuck here
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  2. #2
    Junior Member
    Sep 2012

    Re: Derivatives, Projected object

    If you are looking for the time and velocity with which the object hits the ground you need to set the equation for position equal to zero. You will get two answers, one of them being obviously t=0. Once you solve for the time, then you can plug that value of t into the equation for velocity.
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