Question

A basketball player shoots at a basket 25 ft away. The height of the ball above the ground at time t is given by h=-16t2+32t+6.6. How many seconds after the ball is released does it hit the basket? Hint: When the ball hits the basket h=10ft. Round to the nearest hundredth.

Answer 1

To find the time when the basketball hits the basket, we solve the quadratic equation that represents the height of the ball. After setting the equation equal to 10 feet and using the quadratic formula, we determine the time to be approximately 3.79 seconds.

Step-by-step explanation:

The student is asking about solving a problem related to the projectile motion of a basketball. When the basketball player shoots, the height of the ball is given by the quadratic equation h=-16t2+32t+6.6. We are to find the time when the ball reaches a height of 10 feet, which is the height of the basket.

To solve this problem, we set the equation equal to 10 feet and solve for t, which represents the time in seconds:

10 = -16t2 + 32t + 6.6

Using the quadratic formula, we get two solutions for t. The solution corresponding to the ball hitting the basket on its way down is the larger value of t. After solving, we find that time to be approximately 3.79 seconds, rounded to the nearest hundredth.