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

Question 58886: An object is projected upward from the top of a tower. Its distance in feet above the ground after t seconds is given by s(t)=16t^2+64t+80. How many seconds wil it take to reach ground level?
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
An object is projected upward from the top of a tower. Its distance in feet above the ground after t seconds is given by s(t)=16t^2+64t+80. How many seconds wil it take to reach ground level?
I THINK THE FORMULA IS S = -16T^2+64T+80 WHEN IT IS THROWN UP.
ASSMING SO...
at t=0, when it is thrown up we have
s=0+0+80=80
hence the tower is at 80' from ground.
WHEN THE OBJECTS FALLS ON GROUND ITS DISTANCE FROM GROUND =0
S=0
-16T^2+64T+80=0
T^2-4T-5=0
T^2-5T+T-5=0
T(T-5)+1(T-5)=0
(T-5)(T+1)=0
T=5 SEC.
IT WILL TAKE 5 SEC TO HIT THE GROUND.

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