SOLUTION: An object is projected upward from the top of a water tower. Its distance in feet above the ground after t seconds is given by s(t)=-16t^2+64t+80. How many seconds will it take to

An object is projected upward from the top of a water 
tower. Its distance in feet above the ground after t 
seconds is given by s(t) = -16t^2 + 64t + 80. How 
many seconds will it take to reach ground level? I 
came out with 4 seconds after working it out. thanks 
for checking.

Let's check your answer and see if it is correct:
We substitute 4 seconds into the equation:

                     s(t) = -16t² + 64t + 80              
                     s(4) = -16(4)² + 64(4) + 80                      
                     s(4) = -16(16) + 256 + 80
                     s(4) = -256 + 256 + 80
                     s(4) = 80 feet above ground

After 4 seconds the object is still 80 feet above 
the ground.  So 4 seconds can't be right. Let's
do it right:

Ground level is ground zero, that is, when the
height above the ground s(t) is zero,
So replace s(t) by 0

                     s(t) = -16t² + 64t + 80

                        0 = -16t² + 64t + 80

          16t² - 64t - 80 = 0

Divide every term through by 16

              t² - 4t - 5 = 0

           (t - 5)(t + 1) = 0 

          t - 5 = 0;  t + 1 = 0
              t = 5       t = -1

Answer: 5 seconds.

Discard the negative answer. It takes 5 seconds
to reach the ground.  The water tank is 80 feet high.
At the instant when 2 seconds have passed it reaches
its maximum height of 144 feet (64 feet above the 
water tower). Then it starts falling back down.  At
the instant when 4 seconds have passed, it is back
even with the tower.  Then after 5 seconds have
passed the object reaches the ground.

Edwin
AnlytcPhil@aol.com