A rock is thrown from one side of a river to another. The function h (t) = -16t^2 + 80t + 30 gives the height in inches of the rock t seconds after it has been thrown.

To solve the problems, we will first evaluate \( h(2.5) \).

### a) Find \( h(2.5) \).

Using the given function:

\[
h(t) = -16t^2 + 80t + 30
\]

Substituting \( t = 2.5 \):

\[
h(2.5) = -16(2.5)^2 + 80(2.5) + 30
\]

Calculating \( (2.5)^2 \):

\[
(2.5)^2 = 6.25
\]

So now we substitute:

\[
h(2.5) = -16(6.25) + 80(2.5) + 30
\]

Calculating \( -16(6.25) \):

\[
-16 \times 6.25 = -100
\]

Calculating \( 80(2.5) \):

\[
80 \times 2.5 = 200
\]

Putting it all together:

\[
h(2.5) = -100 + 200 + 30 = 130
\]

Thus,

\[
h(2.5) = 130
\]

### b) What does \( h(2.5) \) represent in the context of this problem?

In the context of this problem, \( h(2.5) \) represents **the height of the rock 2.5 seconds after it has been thrown.**

### c) Does it make sense to find \( h(-10) \) in the context of this problem?

No, it does not make sense because time cannot be negative.

### Summary of Answers:

- a) \( h(2.5) = 130 \)
- b) The height of the rock 2.5 seconds after it has been thrown.
- c) No, it does not make sense because time cannot be negative.