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?

Question

Mathematics

Kash Crane

23 October, 14:11

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?

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Answers (1)

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Roger Lucas · 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