Question 1093601: 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
+ 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 part a, was there a time besides when the ball passed through the basket, when it was 10 feet off the floor? Presents in a graph and table, provide at least 1 analytic (algebraic) explanation.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! I think the formula needed is
-16t^2+19.5t+6.5=f(x)
It passes through 10 feet on its way up and down.
Therefore, set it equal to 10 and when 10 is subtracted from both sides
-16t^2+19.5t-3.5=0
t=-1/32{-19.5+/- sqrt (380.25-224)}; sqrt term is sqrt(156.25)=12.5
t=(-1/32)(-7)=7/32 seconds This is when the ball was on its way to the basket.
t=(-1/32){-19.5-12.5}=1 second
How long would the ball fly?
-16t^2+19.5t+6.5=0. Here, we want the time when h(t)=0.
t=-(1/32)(-19.5+/- sqrt(19.5^2+416); sqrt term=28.22)
t=-(1/32)(47.72)=1.49 sec.
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