Post a New Question


posted by .

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

Answer This Question

First Name
School Subject
Your Answer

Related Questions

More Related Questions

Post a New Question