Let's evaluate the function \( h(t) = -16t^2 + 80t + 30 \).
a) **Finding \( h(2.5) \)**
To find \( h(2.5) \), we can substitute \( t \) with \( 2.5 \):
\[
h(2.5) = -16(2.5)^2 + 80(2.5) + 30
\]
\[
= -16(6.25) + 80(2.5) + 30
\]
\[
= -100 + 200 + 30
\]
\[
= 130
\]
So, \( h(2.5) = 130 \).
**Response area for a)**: 130
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b) **What does \( h(2.5) \) represent in the context of this problem?**
\( h(2.5) \) represents the height of the rock above the ground, measured in inches, at \( t = 2.5 \) seconds after it has been thrown.
**Response area for b)**: The height of the rock above the ground at 2.5 seconds after it has been thrown.
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c) **Does it make sense to find \( h(-10) \) in the context of this problem?**
Finding \( h(-10) \) does not make sense in this context because negative time (before the rock was thrown) is not applicable to the scenario. The function is only relevant for \( t \geq 0 \).
**Response area for c)**: No, it does not make sense to find \( h(-10) \) because negative time is not applicable in this scenario.