April shoots an arrow upward at a speed of 80 feet per second from a distance 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
80 feet
90 feet
125 feet
140 feet

1 answer

To find the maximum height of the arrow described by the quadratic equation \( h = -16t^2 + 80t + 25 \), we need to determine the vertex of the parabola represented by this equation. The vertex formula for the time \( t \) at which the maximum height occurs for a quadratic equation in the form \( at^2 + bt + c \) is given by:

\[ t = -\frac{b}{2a} \]

In this equation:

  • \( a = -16 \)
  • \( b = 80 \)

Plugging in the values:

\[ t = -\frac{80}{2 \times -16} = -\frac{80}{-32} = 2.5 \text{ seconds} \]

Now that we have the time when the arrow reaches its maximum height, we can find the maximum height \( h \) by substituting \( t = 2.5 \) back into the height equation:

\[ h = -16(2.5)^2 + 80(2.5) + 25 \]

Calculating \( (2.5)^2 \) first:

\[ (2.5)^2 = 6.25 \]

So,

\[ h = -16(6.25) + 80(2.5) + 25 \] \[ h = -100 + 200 + 25 \] \[ h = 125 \]

Thus, the maximum height of the arrow is 125 feet.

The answer is: 125 feet.