Question 188763
An object is thrown downward, with an initial speed of 16ft/s, from the top of a building 480 ft. high. How many seconds later will the object hit the ground? Use the equation d = vt + 16t2 , where d is the distance in feet, v is the initial speed, and t is the time in seconds.
----------------------------------------------------
d(t) = vt + 16t^2
16t + 16t^2 = 480
t + t^2 = 3
t^2 + t -3 = 0
t = [-1 +- sqrt(1-4*-3)]/2

t = [-1 +- sqrt(13)]/2
Positive solution:

t = [2.6056]/2
t = 1.3028 seconds
============================
Cheers,
Stan H.