physics/calculus

A stone is thrown straight up from the edge of a roof, 600 feet above the ground, at a speed of 12 feet per second.
A. Remembering that the acceleration due to gravity is -32 feet per second squared, how high is the stone 6 seconds later?
B. At what time does the stone hit the ground?
C. What is the velocity of the stone when it hits the ground?

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  1. h(t)=hi+vi(t)-1/2 * 32*t^2 put t=6 and solve.

    now use the same equation, solve for t when h(t)=0. Ignore the t negative answer.

    vf=vi(t)+gt put in tfinal, and solve, OR
    final KE=initialKE+initialGPE
    or vf^2=vi^2+2(32)(600)

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    bobpursley
  2. h = -16t^2 + 12t + 600

    a) sub in t=6

    b) set -16t^2 + 12t + 600 = 0
    divide by -4
    4t^2 - 3t - 150 = 0

    use the formula to find t = .... (use the positive answer only)

    c) v = -32t + 12
    plug in your answer from b)

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