SOLUTION: if an object is projected upward with an initial velocity off 64 ft per sec from a height of 336 feet then its height t sec after it is projected is defined by the equation h= -16t

Question 449709: if an object is projected upward with an initial velocity off 64 ft per sec from a height of 336 feet then its height t sec after it is projected is defined by the equation h= -16t^2 + 64 t + 336. how many sec after it is projected will it hit the ground?
Found 2 solutions by jorel1380, lwsshak3:
Answer by jorel1380(3719) (Show Source): You can put this solution on YOUR website!
0=-16t2+64t+336
16t2-64t-336=0
t2-4t-21=0
(t-7)(t+3)=0
t=7 or -3
Throwing out the negative answer, we get the time t to be 7 seconds..
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
if an object is projected upward with an initial velocity off 64 ft per sec from a height of 336 feet then its height t sec after it is projected is defined by the equation h= -16t^2 + 64 t + 336. how many sec after it is projected will it hit the ground?
..
When object hits the ground its height will be=0
0= -16t^2 + 64 t + 336
-16t^2 + 64 t + 336=0
divide by -16
t^2-4t-21=0
(t-7)(t+3)=0
t-7=0
t=7 seconds
t+3=0
t=-3 (reject, t≥0)
ans:
object will hit the ground 7 seconds after it is projected.
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