Question
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
h = -16t^2 + 80t + 25
80 feet
90 feet
125 feet
140 feet
Answers
GPT-4o mini
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**.
\[
t = -\frac{b}{2a}
\]
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**.