Question 45
April shoots an arrow upward at a speed of 80 feet per second from a platform 25 feet high. The pathway of the arrow can be represented by the equation below, where "h" is the height and "t" is the time in seconds. What is the maximum height of the arrow?
h = -16t^2 + 80t + 25
To find the maximum height of the arrow, we need to determine the vertex of the quadratic equation h = -16t^2 + 80t + 25.
The vertex of a quadratic equation in the form h = at^2 + bt + c is given by the formula t = -b / (2a).
In this case, a = -16 and b = 80. Plugging these values into the formula, we get:
t = -80 / (2*(-16))
t = -80 / (-32)
t = 2.5
Now, plug t = 2.5 back into the equation h = -16t^2 + 80t + 25 to find the maximum height:
h = -16(2.5)^2 + 80(2.5) + 25
h = -16(6.25) + 200 + 25
h = -100 + 225
h = 125
Therefore, the maximum height of the arrow is 125 feet.