An arrow is shot with an initial upward velocity of 150 feet per second from a height of 5 feet above the ground. The equation h=−16t2+150t+5 models the height in feet t seconds after the arrow is shot. After the arrow passes its maximum height, it comes down and hits a target that was placed 20 feet above the ground. About how long after the arrow was shot does it hit its intended target?
+37159 Calculate h = 20
20 = -16t^2 + 150t + 5 solve for t using Quadratic Formula
you will get two values....one is arrow going up....the other (larger) is arrow comong down...
