Applied Calculus

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Suppose that an object is thrown down towards the ground from the top of the 1100-ft tall Sears
Tower with an initial velocity of – 20 feet per second (the negative velocity indicates that the object
is moving towards the ground). Given that the acceleration due to gravity of the object is a constant –
32 feet per second per second, determine how far above the ground the object is exactly six seconds
after being thrown.

  • Applied Calculus -

    h = Hi + Vi t + (1/2) a t^2
    Hi = 1100
    Vi = - 20
    a = -32
    so
    h = 1100 -20 t - 16 t^2
    if t=6
    h = 1100 -20*6 - 16*36

    = 404 ft

  • Applied Calculus -

    all this from integrating acceleration
    a = -32 ft/s^2
    so
    dv/dt = -32
    v = Vi - 32 t

    dh/dt = v = Vi - 32 t
    h = Hi + Vi t - (1/2)(32)t^2

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