Question

A ball is launched upward at 64 feet per second from a platform that is 36 feet high.

The function that models this situation is h(t)=−16t2+64t+36
where t is the time in seconds and h(t)
is the height of the ball.

Find the maximum height the ball reaches and how long it will take to get there.

(1 point)
A) What key feature of the graph is the problem asking you to find?
(answer choices: the vertex, y-intercept, x-intercept)

B) The ball will reach a maximum height of
feet in
seconds.

Answer 1

To solve this problem, we need to analyze the quadratic function given by \( h(t) = -16t^2 + 64t + 36 \).

### A) What key feature of the graph is the problem asking you to find?
The problem is asking for the maximum height of the ball, which corresponds to the vertex of the parabola represented by the function. Since the parabola opens downward (as indicated by the negative coefficient of \( t^2 \)), the vertex will represent the maximum point.

**Answer:** The vertex

### B) The ball will reach a maximum height of _____ feet in _____ seconds.

To find the vertex of the quadratic function, we can use the formula for the time \( t \) at which the maximum height occurs in a quadratic equation \( ax^2 + bx + c \):

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

Here, \( a = -16 \) and \( b = 64 \).

Calculating \( t \):

\[
t = \frac{-64}{2 \times -16} = \frac{-64}{-32} = 2
\]

Now, we will substitute \( t = 2 \) back into the function \( h(t) \) to find the maximum height:

\[
h(2) = -16(2^2) + 64(2) + 36
\]
\[
= -16(4) + 128 + 36
\]
\[
= -64 + 128 + 36
\]
\[
= 100
\]

Thus, the maximum height of the ball is 100 feet, and it reaches this height in 2 seconds.

**Answer:** The ball will reach a maximum height of **100** feet in **2** seconds.