Question

A basketball is shot from the free throw line.
H(t)=-16t^2 +40t+6 gives the height h(t) of thw basketball in feet (t) seconds after it is released.

What is the value of h(1)?
What does h(2.5) represent in the context of this problem?

Answer 1

To find \( h(1) \), we can substitute \( t = 1 \) into the function \( H(t) = -16t^2 + 40t + 6 \).

Calculating \( h(1) \):

\[
h(1) = -16(1)^2 + 40(1) + 6
\]
\[
h(1) = -16(1) + 40 + 6
\]
\[
h(1) = -16 + 40 + 6
\]
\[
h(1) = 24
\]

So, \( h(1) = 24 \) feet. This means that 1 second after the basketball is released, its height is 24 feet.

Next, to interpret \( h(2.5) \) in the context of the problem, we can find its value by substituting \( t = 2.5 \) into the function \( H(t) \):

Calculating \( h(2.5) \):

\[
h(2.5) = -16(2.5)^2 + 40(2.5) + 6
\]
\[
h(2.5) = -16(6.25) + 100 + 6
\]
\[
h(2.5) = -100 + 100 + 6
\]
\[
h(2.5) = 6
\]

So, \( h(2.5) = 6 \) feet. In the context of this problem, this means that 2.5 seconds after the basketball is released, its height is 6 feet. This indicates that the ball is on its way down after reaching its peak height.