Certainly! Let’s calculate the height of the basketball after 1.5 seconds using the given equation:
[tex]\[ h(t) = -16t^2 + 25t + 6 \][/tex]
We need to substitute [tex]\( t = 1.5 \)[/tex] seconds into the equation to find the height.
1. Plug [tex]\( t = 1.5 \)[/tex] into the height equation:
[tex]\[
h(1.5) = -16 \cdot (1.5)^2 + 25 \cdot 1.5 + 6
\][/tex]
2. First, calculate [tex]\( (1.5)^2 \)[/tex]:
[tex]\[
(1.5)^2 = 2.25
\][/tex]
3. Multiply by [tex]\(-16\)[/tex]:
[tex]\[
-16 \cdot 2.25 = -36
\][/tex]
4. Multiply [tex]\( 25 \times 1.5 \)[/tex]:
[tex]\[
25 \cdot 1.5 = 37.5
\][/tex]
5. Now, add these values along with the initial height of 6 feet:
[tex]\[
h(1.5) = -36 + 37.5 + 6
\][/tex]
6. Perform the addition:
[tex]\[
-36 + 37.5 = 1.5 \\
1.5 + 6 = 7.5
\][/tex]
So, after performing these calculations, we find that the height of the basketball after 1.5 seconds is:
[tex]\[ h = 7.5 \][/tex]
Since the problem asks for the height rounded to the nearest tenth of a foot, and 7.5 is already rounded to the nearest tenth, the basketball is:
[tex]\[
\boxed{7.5}
\][/tex]
Hence, after 1.5 seconds, the basketball is 7.5 feet above the ground.
