An arrow is shot with an initial upward velocity of 100 feet per second

Question

An arrow is shot with an initial upward velocity of 100 feet per second from a height of 5 feet above the ground. The equation h= -16t^2 + 100t + 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?



Answer 1

So this question is basically asking you to find where 20= -16t^2+100t+5 because you want to find the time for how long after the arrow was launched does it hit the intended target of 20 feet. Using a graphing calculator you put -16t^2+100t+5 into y1 and 20 into y2. Once you do that you can find the intersection point and the x coordinate is your answer because the x value represents the time. So the answer is 6.096 seconds. If you need an exact answer then you need to use the quadratic formula and you would get (25+the square root of 565)/ 8