Question

Sean is teaching his little brother Victor to play basketball. From near the net, Victor throws the ball up in the air from a height of 5 feet and with a velocity of 22 feet per second. On the way down, the ball goes straight through the hoop at a height of 10 feet above the ground. Which equation can you use to find how many seconds the ball is in the air before going through the hoop? If an object travels upward at a velocity of v feet per second from s feet above the ground, the object's height in feet, h, a ter t seconds can be modeled by the formula

h=-16t²+vt+s. 10=-16t²+22t+5
10=-16t²+5t+22
To the nearest tenth of a second, how long is the ball in the air before going through the hoop?

Answer 1

To find the duration the basketball is in the air before passing through the hoop, we used the motion equation 10 = -16t² + 22t + 5, which led us to the correct time of 3.79 seconds.

Step-by-step explanation:

To solve for the time that the basketball remains in the air until it passes through the hoop, we can use a quadratic equation that describes the motion of the ball. The equation is of the form h = -16t² + vt + s, where h is the final height of the ball, v is the initial velocity, s is the initial height, and t is the time. Plugging the given values into the equation with h equal to 10 feet (the height of the hoop), we get 10 = -16t² + 22t + 5.

To find the time t, we adjust the equation to 0 = -16t² + 22t - 5 and apply the quadratic formula or factor where applicable. We find that the ball is in the air for two possible times, but we are interested in the longer duration, which represents the ball passing through the hoop on its way down. Using the formula yields t = 3.79 seconds as the time the ball is in the air before going through the hoop to the nearest tenth of a second.