Question

A basketball is thrown with an initial upward velocity of 30 feet per second from a height of 6 feet above the ground. The equation h=-16t^(2)+30t+6 models the height in feet t seconds after the basketball 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

For this case we have the following equation:
h = -16t ^ (2) + 30t + 6
We substitute the value of h = 10 in the equation:
10 = -16t ^ (2) + 30t + 6
Rewriting we have:
0 = -16t ^ (2) + 30t + 6-10
0 = - 16t ^ (2) + 30t - 4
We look for the roots of the polynomial:
t1 = 0.144463904
t2 = 1.730536096
"the ball passes its maximum height, it comes down and then goes into the hoop". Therefore, the correct root is:
t = 1.730536096
Answer:
It goes into the hoop after:
t = 1.73 seconds