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
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.
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.