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?
a. 0.08 seconds
b. 1.56 seconds
c. 1.48 seconds
d. 1.83 seconds
To find out when the ball goes into the hoop, we need to solve for t in the equation h = -16t^2 + 25t + 8 when h = 10.
Substituting h = 10 into the equation, we have:
10 = -16t^2 + 25t + 8
Rearranging the equation to standard form, we get:
-16t^2 + 25t - 2 = 0
To solve this quadratic equation, we can use the quadratic formula, which states that for an equation of the form ax^2 + bx + c = 0, the solutions are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -16, b = 25, and c = -2. Substituting these values into the quadratic formula, we have:
t = (-25 ± √(25^2 - 4(-16)(-2))) / (2(-16))
Simplifying further, we get:
t = (-25 ± √(625 - 128)) / (-32)
t = (-25 ± √497) / (-32)
Calculating the square root of 497, we get:
t = (-25 ± 22.31) / (-32)
So the two possible values for t are:
t = (-25 + 22.31) / (-32) ≈ 0.08
t = (-25 - 22.31) / (-32) ≈ 1.56
Since the time cannot be negative, the ball goes into the hoop approximately 1.56 seconds after it was thrown.
Therefore, the correct option is:
b. 1.56 seconds