Calculus

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A ball is thrown straight down from the top of a 220-foot building with an initial velocity of -22 feet per second. What is its velocity after 3 seconds? What is its velocity after falling 108 feet?

  • Calculus -

    equation:
    s = -16t^2 - 22t + 220
    v = ds/dt = -32t - 22
    so at t=3
    v = -16(3) - 22 = -70 ft/sec

    when s = 108
    108 = -16t^2 - 22t + 220
    16t^2 + 22t - 112 = 0
    t = (-22 ± √7652)/32 = 2.046 sec or a negative time

    so when t = 2.046
    v = -16(2.046) - 22 = - 54 ft/s

  • Calculus -

    s = -16t^2 - 22t + 220
    v = ds/dt = -32t - 22
    so at t=3
    v = -32(3) - 22 = -118 ft/sec

    when s = 108
    108 = -16t^2 - 22t + 220
    16t^2 + 22t - 112 = 0
    t = (-22 ± √7652)/32 = 2.046

    so when t = 2.046
    v = -32(2.046) - 22 = -87.472 ft/s


    The process is correct but the t=2.046 should be substituted to the first derivative which is velocity function.

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