To find \( h(2.5) \), we can substitute \( t = 2.5 \) into the function \( h(t) = -16t^2 + 80t + 30 \).
Calculating \( h(2.5) \):
\[ h(2.5) = -16(2.5)^2 + 80(2.5) + 30 \]
First, calculate \( (2.5)^2 \):
\[ (2.5)^2 = 6.25 \]
Now substitute this value into the equation:
\[ h(2.5) = -16(6.25) + 80(2.5) + 30 \]
Calculating each term:
\[ -16(6.25) = -100 \] \[ 80(2.5) = 200 \]
Combining all the terms:
\[ h(2.5) = -100 + 200 + 30 \] \[ h(2.5) = 100 + 30 = 130 \]
Thus, \( h(2.5) = 130 \).
Interpretations:
In the context of this problem, \( h(2.5) = 130 \) means that 2.5 seconds after the rock was thrown, its height above the ground is 130 inches.
About \( h(-10) \):
Now, regarding \( h(-10) \):
Finding \( h(-10) \) would represent the height of the rock at -10 seconds, which does not make sense in this context since time cannot be negative in this scenario. Therefore, it is not meaningful to find \( h(-10) \) as it does not correspond to a real point in time after the rock has been thrown.