Question

A ball is launched upward at 48 feet per second from a platform that is 100 feet high. The function that models this situation is h(t)=−16t2+48t +100
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.

What key feature of the graph is the problem asking you to find?



(1 point)
Responses

the x intercept
the x intercept

the vertex
the vertex

The y-intercept
The y-intercept

the rate of change
the rate of change

the axis of symmetry
the axis of symmetry

Answer 1

To find the maximum height the ball reaches and the time it will take to get there, we need to determine the vertex of the quadratic function \( h(t) = -16t^2 + 48t + 100 \).

The vertex form of a quadratic function \( ax^2 + bx + c \) has its vertex (which gives the maximum height for a downward-opening parabola) at the time \( t \) given by:

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

In this case, \( a = -16 \) and \( b = 48 \).

Plugging in these values:

\[
t = -\frac{48}{2 \times -16} = -\frac{48}{-32} = 1.5 \text{ seconds}
\]

Now, to find the maximum height, we substitute \( t = 1.5 \) back into the height function \( h(t) \):

\[
h(1.5) = -16(1.5)^2 + 48(1.5) + 100
\]

Calculating \( (1.5)^2 \):

\[
(1.5)^2 = 2.25
\]

Now substituting:

\[
h(1.5) = -16(2.25) + 48(1.5) + 100
\]
\[
h(1.5) = -36 + 72 + 100
\]
\[
h(1.5) = 136 \text{ feet}
\]

Thus, the maximum height the ball reaches is **136 feet** and it takes **1.5 seconds** to get there.

### Summary:

- **Maximum Height:** 136 feet
- **Time to reach maximum height:** 1.5 seconds

As for the key feature of the graph that the problem is asking for, the answer is **the vertex**.