Question 47975: A 5.6 feet tall woman is shooting a free throw. The path of the basketball is parabolic in shape and the ball reaches its maximum height of 11.5 feet when the ball is 10 feet from the player. (i) Find the equation for the path of the ball. Let x be the horizontal distance from the shooter, and y be the height of the ball. (ii) the ball hits the front of the rim, which is 10 feet high. How far is the shooter from the rim?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A 5.6 feet tall woman is shooting a free throw. The path of the basketball is parabolic in shape and the ball reaches its maximum height of 11.5 feet when the ball is 10 feet from the player. (i) Find the equation for the path of the ball. Let x be the horizontal distance from the shooter, and y be the height of the ball. (ii) the ball hits the front of the rim, which is 10 feet high. How far is the shooter from the rim?
Draw the picture. The ball starts at (0,5.6), rises to (10,11.5), and
decends to (x,10).
The equation form is ax^2+bx+c=y
Substituting (0,5.6) you get c=5.6
Substituting (10,11.5) you get 100a+10b+5.6=11.5
Then 100a + 10b = 5.9
Since (10,11.5) is a maximum point, -b/2a = 10; so b=-20a
Substituting that value for "b" into 100a+10b=5.9 you get:
100a+10(-20a)=5.9
-100a=5.9
a=-0.059
Substituting that into b=-20a you get:
b=-20(-0.059)=1.18
So the equation is y=-0.059x^2+1.18x+5.6
Substituting y=11.5 into this equation you get x=approximately 4.96 ft.
Cheers,
Stan H.
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