To solve the problems, we will evaluate the function \( h(t) = -16t^2 + 40t + 6 \).
1. What is the value of \( h(1) \)?
We substitute \( t = 1 \) into the function:
\[ h(1) = -16(1)^2 + 40(1) + 6 \] \[ = -16(1) + 40 + 6 \] \[ = -16 + 40 + 6 \] \[ = 24 \]
So, \( h(1) = 24 \) feet.
2. What does \( h(2.5) \) represent in the context of this problem?
To find out what \( h(2.5) \) represents, we calculate \( h(2.5) \):
\[ h(2.5) = -16(2.5)^2 + 40(2.5) + 6 \] \[ = -16(6.25) + 100 + 6 \] \[ = -100 + 100 + 6 \] \[ = 6 \]
Thus, \( h(2.5) = 6 \) feet.
In the context of the problem, \( h(2.5) \) represents the height of the basketball 2.5 seconds after it is released. The height is 6 feet at that time.
3. What does \( h(-1) \) represent in the context of this problem?
To interpret \( h(-1) \), we could calculate it as follows:
\[ h(-1) = -16(-1)^2 + 40(-1) + 6 \] \[ = -16(1) - 40 + 6 \] \[ = -16 - 40 + 6 \] \[ = -50 \]
So, \( h(-1) = -50 \) feet.
In the context of this problem, \( h(-1) \) would represent the height of the basketball at a time that is 1 second before it is released. Since time cannot physically be negative in this scenario, it indicates a hypothetical situation before the ball was released and the negative height values don't have a practical meaning in this context; they suggest that the ball would be below ground level (which isn't possible in a practical setting). Thus, it indicates a point in time before the shot was taken, but it doesn't reflect any physical reality related to the ball's actual trajectory.
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