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22 October, 23:18

A ball is thrown vertically upward from the top of a building 64 feet tall with an initial velocity of 48 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s equals 64 plus 48 t minus 16 t squared. (a) After how many seconds does the ball strike the ground? (b) After how many seconds will the ball pass the top of the building on its way down?

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  1. 23 October, 02:55
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    a) t = 4 secs

    b) t = 3 secs

    Step-by-step explanation:

    h = 64 ft

    Initial velocity = 48 ft/s

    S = 64 + 48t - 16t^2

    a) Set S = 0 and solve for t

    0 = 64 + 48t - 16t^2

    Divide through by 16

    0 = 4 + 3t - t^2

    t^2 - 3t - 4 = 0

    t^2 - 4t + t - 4 = 0

    t (t - 4) + 1 (t - 4) = 0

    (t + 1) (t - 4) = 0

    t + 1 = 0 or t - 4 = 0

    t = - 1 or t = 4

    t = {-1,4}

    t = 4 seconds

    b) Set S = 64 and solve for t

    64 = 64 + 48t - 16t^2

    64 - 64 = 48t - 16t^2

    0 = 48t - 16t^2

    Divide through by 16

    0 = 3t - t^2

    t^2 - 3t = 0

    t (t - 3) = 0

    t = 0 or t-3 = 0

    t = 0 or t = 3

    t = {0,3}

    t = 3 secs
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