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A baseball is thrown upward from ground level with an initial velocity of 48 feet per second, and its height h in feet is given by , where t is the time in seconds.
At what time does the ball reach 10 feet? Round to the nearest tenth.

Step by step please

  • Math -

    h(t) = 48t - 16t^2
    so, plug in 10 for h and solve for t:

    10 = 48t - 16t^2
    8t^2 - 24t + 5 = 0
    t = (6±√26)/4
    or
    t=0.225s (on the way up)
    t=2.774s (on the way back down)

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