A basketball is thrown with an initial upward velocity of 25 feet per second from a height of 8 feet above the ground. The equation h = -16t^2 + 25t + 8 models the height in feet t seconds after it is thrown. After the ball passes its maximum height, it comes down and then goes into the hoop at a height of 10 feet above the ground. About how long after it was thrown does it go into the hoop?

Answer 1

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.

Answer 2

wrong again
-16t^2 + 25t + 8 = 10
-16t^2 + 25t - 2 = 0
t = (25±√497)/32
That is, t ≈ 1.478 seconds

Answer 3

Apologies for the mistake in my previous response. You are correct.

By solving the quadratic equation -16t^2 + 25t - 2 = 0 using the quadratic formula, we get:

t = (25 ± √497)/32

Using a calculator to evaluate this expression, we find that t is approximately 1.478 seconds.

Therefore, approximately 1.478 seconds after it was thrown, the ball goes into the hoop.