Pre-Calc

ball thrown vertically upward with an initial velocity of 80 ft per second. The distance s(in ft) of the ball from the ground after t seconds is s=80t-16t^2.
a. draw the illustration.
b. for what time interval is ball more than 96 ft above ground?
c. what's the max height of ball?
d. after how many seconds does the ball reach max height?

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asked by Andy G
  1. b. d = 80t - 16t^2 = 96

    -16t^2 + 80t - 96 = 0
    Divide both sides by 16:
    -t^2 + 5t - 6 = 0
    Use Quadratic Formula and get:
    t = 2s, and 3 s.
    Interval: 2< T <3.

    c. h = (V^2-Vo^2)/2g
    h = (0-80^2)/64 = 100 Ft.

    d. V = Vo + g*t = 0 at max ht.
    t = -Vo/g = -80/-32 = 2.50 s.

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    posted by Henry

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