To find out when the basketball will hit the ground, we need to determine when the height \( h \) equals 0. The height of the basketball is given by the equation:
\[ h(t) = -16t^2 + 20t + 64 \]
We set \( h(t) \) to 0:
\[ -16t^2 + 20t + 64 = 0 \]
To solve this quadratic equation, we can use the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = -16 \), \( b = 20 \), and \( c = 64 \).
First, we calculate the discriminant:
\[ b^2 - 4ac = 20^2 - 4(-16)(64) \] \[ = 400 + 4096 \] \[ = 4496 \]
Now, we can substitute the values into the quadratic formula:
\[ t = \frac{-20 \pm \sqrt{4496}}{2(-16)} \]
Calculating \( \sqrt{4496} \):
\[ \sqrt{4496} \approx 67.1 \]
Now substitute this back in:
\[ t = \frac{-20 \pm 67.1}{-32} \]
Calculating the two possible solutions for \( t \):
-
Using the positive root: \[ t = \frac{-20 + 67.1}{-32} = \frac{47.1}{-32} \approx -1.47 \quad \text{(not valid since time cannot be negative)} \]
-
Using the negative root: \[ t = \frac{-20 - 67.1}{-32} = \frac{-87.1}{-32} \approx 2.72 \]
Therefore, the basketball will hit the ground approximately \( t \approx 2.72 \) seconds after it is shot.
1 answer