To determine when the basketball will hit the ground, we need to set the height \(h\) to 0 in the equation:
\[
h = -16t^2 + 20t + 64
\]
Setting \(h\) to 0 gives us:
\[
0 = -16t^2 + 20t + 64
\]
We can rearrange the equation:
\[
-16t^2 + 20t + 64 = 0
\]
To make calculations easier, we can multiply the entire equation by -1:
\[
16t^2 - 20t - 64 = 0
\]
Next, we can either use the quadratic formula or factor it. Let's try factoring first. We look for two numbers that multiply to \(16 \cdot -64 = -1024\) and add to \(-20\).
However, it might be easier to use the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
In our case, \(a = 16\), \(b = -20\), and \(c = -64\).
Plugging in these values:
\[
t = \frac{-(-20) \pm \sqrt{(-20)^2 - 4 \cdot 16 \cdot (-64)}}{2 \cdot 16}
\]
This simplifies to:
\[
t = \frac{20 \pm \sqrt{400 + 4096}}{32}
\]
Calculating the square root:
\[
t = \frac{20 \pm \sqrt{4496}}{32}
\]
Now, let's simplify \(\sqrt{4496}\):
\[
\sqrt{4496} \approx 67.07
\]
Substituting back gives:
\[
t = \frac{20 \pm 67.07}{32}
\]
Calculating the two possible values for \(t\):
-
Using the positive root:
\[
t = \frac{20 + 67.07}{32} \approx \frac{87.07}{32} \approx 2.72 \text{ seconds}
\]
-
Using the negative root:
\[
t = \frac{20 - 67.07}{32} \approx \frac{-47.07}{32} \approx -1.47 \text{ seconds (not possible)}
\]
Since time cannot be negative, we discard the second value.
Thus, the basketball will hit the ground approximately 2.72 seconds after being shot.
So the final answer is:
\[
\boxed{2.72 \text{ seconds}}
\]
