Calculus

A ball is shot straight up into the air with initial velocity of 47 ft/sec. Assuming that the air resistance can be ignored, how high does it go?

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asked by Anonymous
  1. A ball is shot straight up into the air with initial velocity of 47 ft/sec

    Free fall, s in ft, t in sec
    s = s0 + v0t - 16t^2
    height, s0 = 0
    velocity, v0 = 47
    s = 0 + 47t - 16t^2
    s = 47t - 16t^2

    Instantaneous velocity, v
    v = ds/dt = v0 - 32t
    v = 47 ft/sec
    v = 47 - 32t

    To find time, t
    At maximum value or turning point, v = 0
    v = 47 - 32t
    0 = 47 - 32t
    32t = 47
    t = 47/32 = 1.47 sec to reach max height

    s = 47t - 16t^2
    t = 1.47 sec
    s = 47(1.47) - 16(1.47)^2
    s = 69.09 - 16(2.16)
    s = 69.09 - 34.56
    s = 34.53 feet

    So, it takes 1.47 sec to reach its maximum height of 34.53 feet

    Check my math

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    posted by helper

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