A basketball is thrown with an initial upward velocity of 30 feet per second from a height of 6 feet above the ground. The equat
ion 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?
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
