Question 394787
A ball is thrown vertically upward from the top of a building 32 feet tall with an initial velocity of 16 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s=32+16t-16t^2. 
a. after how many seconds does the ball strike the ground? 
set s to zero and solve for t:
s=32+16t-16t^2
0=32+16t-16t^2
0=2+t-t^2
t^2-t-2=0
(t-2)(+1) = 0
t = {-1, 2}
the -1 is an extraneous solution leaving us with:
t = 2 seconds
.
b. after how many seconds does the ball pass the top of the building on its way down?
set s = 32 and solve for t:
s=32+16t-16t^2
32=32+16t-16t^2
0=16t-16t^2
0=t-t^2
0=t(1-t)
t = {0,1}
the 0 represents when it was first thrown
so, 1 second is your answer