SOLUTION: Margaret shoots an arrow into the air. The equation for the height (in feet) of the tip of the arrow is h=8+64t−16t^2 To find the time at which the arrow is 56 feet above the

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Margaret shoots an arrow into the air. The equation for the height (in feet) of the tip of the arrow is h=8+64t−16t^2 To find the time at which the arrow is 56 feet above the      Log On


   



Question 869239: Margaret shoots an arrow into the air. The equation for the height (in feet) of the tip of the arrow is h=8+64t−16t^2 To find the time at which the arrow is 56 feet above the ground, we replace h with 56 obtain
56=8+64t−16t^2


Solve this equation for t to find the times at which the arrow is 56 feet above the ground.
The arrow is 56 feet above the ground after_______ seconds.

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Margaret shoots an arrow into the air. The equation for the height (in feet) of the tip of the arrow is h=8+64t−16t^2 To find the time at which the arrow is 56 feet above the ground, we replace h with 56 obtain
56=8+64t−16t^2
-16t^2+64t-48=0
-t^2+4t-3=0
t^2-4t+3=0
(t-1)(t-3)=0
t=1
or
t=3
the arrow is 56 feet above the ground after 1 sec on the way up and after 3 sec on the way down