Question

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?
*
1 point
h = -16t^2 + 80t + 25
80 feet
90 feet
125 feet
140 feet

Answer 1

To find the maximum height of the arrow, we can use the given quadratic equation:

\[ h = -16t^2 + 80t + 25 \]

This equation is in the standard form \( h = at^2 + bt + c \), where \( a = -16 \), \( b = 80 \), and \( c = 25 \).

The maximum height of a quadratic function in this form occurs at the vertex, which can be found using the formula:

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

Substituting the values of \( a \) and \( b \):

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

Now, we can substitute \( t = 2.5 \) back into the height equation to find the maximum height \( h \):

\[ h = -16(2.5)^2 + 80(2.5) + 25 \]
\[ h = -16(6.25) + 200 + 25 \]
\[ h = -100 + 200 + 25 \]
\[ h = 125 \]

Thus, the maximum height of the arrow is:

**125 feet.**