Question

What path might a basketball travel if you shot from the foul line to a 10 foot hoop?

Steve races down the basketball court, then stops and shoots -- HE SCORES!!! The ball travels through the air and through the basket (10 feet off the ground) according to the function

h = -16t2+ 19.5t + 6.5, where h represents the height of the basketball above the floor, and t represents the number of seconds that the ball was in the air. How long did it take for the ball to pass through the basket?

The next time down the court, Steve took another shot, shooting the ball according to the same function as in part a, but since he was a few feet farther from the basket, the shot was an Air Ball (missing everything completely). How long was the ball in the air before it hit the floor?

Returning to the first part, was there a time, besides when the ball passed through the basket, when it was 10 feet off the floor? Show the results in a graph and a table, and provide at least 1 analytic (algebraic) explanation.

Answer 1

The ball travels through the air according to the function h = -16t²+ 19.5t + 6.5, where h represents the height of the basketball above the floor, and t represents the number of seconds that the ball was in the air. The basket is 10 ft. off the ground, so that the time taken by the ball to pass through the basket will be given by 10 =-16t²+19.5t+6.5 or, -16t²+ 19.5t - 3.5 = 0 or, 16t²- 19.5t +3.5 = 0 or, 32t² -39t +7 = 0. On using the quadratic formula, we have t = [39±√{ (-39)² -4*32*7}]/2*32 = [39±√(1521-896)]/64 = (39±√625)/2 = (39±25)/64. The path of the ball is a parabola so that the ball will rise to a height higher than 10 ft. before coming down and passing through the basket. Therefore, we will ignore the (-)ve sign. Thus, t = (39+25)/64 = 64/64 = 1 second .

When the ball misses the basket, let the time it was it the air be 10 seconds. Then -16t²+19.5t+ 6.5=0 or, 32t²-39t -13= 0. On using the quadratic formula, we have t=                   [39±√{(-39)² -4*32*(-13)}]/2*32 = [39±√(1521+1664)]64 = (39±√3185)/64 = (39±56.44)/64 . Now, since t cannot be negative, hence t = (39+56.44)/64 = 95.44/64 = 1.49 seconds (on rounding off to 2 decimal places).

As mentioned earlier, the path of the ball is a parabola so that the ball will rise to a height higher than 10 ft. before coming down and passing through the basket. When the ball was 10 ft. high on the way to the basket, we have t = (39-25)/64 = 14/64 = 0.22 second (on rounding off to 2 decimal places).

A graph is attached.