To find the time when the ball goes into the hoop, we need to solve the equation h = 10 for t.
-16t^2 + 25t + 8 = 10
Rearranging the equation, we get:
-16t^2 + 25t - 2 = 0
Using the quadratic formula t = (-b ± sqrt(b^2 - 4ac))/(2a), where a = -16, b = 25, and c = -2, we can solve for t.
t = (-25 ± sqrt(25^2 - 4(-16)(-2)))/(2(-16))
Simplifying the equation, we get:
t = (-25 ± sqrt(625 - 128))/(-32)
t = (-25 ± sqrt(497))/(-32)
Since time cannot be negative in this context, we can ignore the negative value of t.
t ≈ (-25 + sqrt(497))/(-32)
Using a calculator, we find t ≈ 0.209.
So, about 0.209 seconds after it was thrown, the ball goes into the hoop.