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A ball is dropped from the top of a building. The height, , of the ball above the ground (in feet) is given as a function of time, , (in seconds) by
y = 1640 - 16t^2
y'= -32t
When does the ball hit the ground, and how fast is it going at that time? Give your answer in feet per second and in miles per hour (1 ft/sec = 15/22 mph).

v(t) = mph
v(t) = ft per second

  • calculus - ,

    for the first part solve
    0 = 1640 - 16t^2
    16t^2 = 1640
    t^2 = 102.5
    t = 10.124 seconds

    sub that into y' to get the velocity in ft/sec
    Use the formula given to you to change to mph

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