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