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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?

Respuesta :

carlosego
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

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