To determine when the basketball hits the ground, we need to find the time \( t \) when the height \( h \) is zero. The height equation is given by:
\[ h = -16t^2 + 20t + 64 \]
We set \( h \) equal to zero and solve for \( t \):
\[ 0 = -16t^2 + 20t + 64 \]
This is a quadratic equation in the standard form \( at^2 + bt + c = 0 \), where:
- \( a = -16 \)
- \( b = 20 \)
- \( c = 64 \)
We can solve this using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ t = \frac{-20 \pm \sqrt{20^2 - 4(-16)(64)}}{2(-16)} \]
Calculating the discriminant:
\[ 20^2 = 400 \] \[ 4(-16)(64) = -4096 \implies \text{Since it's negative, we actually do: } 4(16)(64) = 4096 \] \[ b^2 - 4ac = 400 + 4096 = 4496 \]
Now we plug this back into the formula:
\[ t = \frac{-20 \pm \sqrt{4496}}{-32} \]
Calculating \( \sqrt{4496} \):
\[ \sqrt{4496} \approx 67.07 \]
Now plug this back in:
\[ t = \frac{-20 \pm 67.07}{-32} \]
Calculating the two potential times:
-
Using the positive root: \[ t = \frac{-20 + 67.07}{-32} = \frac{47.07}{-32} \approx -1.47 \text{ (Not physically relevant, as time cannot be negative)} \]
-
Using the negative root: \[ t = \frac{-20 - 67.07}{-32} = \frac{-87.07}{-32} \approx 2.72 \]
Thus, the basketball will hit the ground approximately at:
\[ \boxed{2.72} \text{ seconds} \]