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Question 521058: The equation y = 0.5x - .01x^2 represents the parabolic flight of a certain cannonball shot at an angle of 26 degrees with the horizon and an initial velocity of 25 meters per second. In this equation, y is the height of the cannonball, in meters, and the x is the vertical distance traveled, in meters.
First, Given the points (10,4) abd (40,0) lie on the parabola, at what x-coordinate must the vertex lie?
- I figured out that it is 25 meters because y = -0.01x^2 + 0.5x. a= -0.01 and b= 0.5. The I plugged it into -b/2a and got x=25.
Secondly, use the equation and your answer to part one to find the maximum heigth of the baseball.
- I plugged it into the equation of y = -0.01x^2 + 0.5x. It looked like this
y = -0.01(25)^2 + 0.5(25)
y= 6.25
Now, it says to use the point (0,0) and the location of the vertex to find the total horizontal distance that the baseball will travel. Im not sure how to find that. Could you help me?
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The equation y = 0.5x - .01x^2 represents the parabolic flight of a certain cannonball shot at an angle of 26 degrees with the horizon and an initial velocity of 25 meters per second. In this equation, y is the height of the cannonball, in meters, and the x is the vertical distance traveled, in meters.
First, Given the points (10,4) abd (40,0) lie on the parabola, at what x-coordinate must the vertex lie?
- I figured out that it is 25 meters because y = -0.01x^2 + 0.5x. a= -0.01 and b= 0.5. The I plugged it into -b/2a and got x=25.
Secondly, use the equation and your answer to part one to find the maximum heigth of the baseball.
- I plugged it into the equation of y = -0.01x^2 + 0.5x. It looked like this
y = -0.01(25)^2 + 0.5(25)
y= 6.25
Now, it says to use the point (0,0) and the location of the vertex to find the total horizontal distance that the baseball will travel. Im not sure how to find that.
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If the cannonball starts at (0,0) and has gone a horizontal distance of 25
by the time it reaches its vertex, it will go another 25 by the time
it hits the ground.
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Cheers,
Stan H.
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Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
First let's fix a couple of errors. In the first place,  doesn't represent the vertical distance traveled.  is the height and the height is on the vertical axis. represents the horizontal distance traveled. Second, the point ) is on the parabola but the point ) most definately is NOT. ) is on the parabola because \ -\ 0.01(40^2)\ =\ 20\ -\ 16\ =\ 4) .
The rest is correct.
Now, for the last bit. Parabolas are symmetrical about the axis which is a line through the vertex. In this case, since the parabola is concave down, the axis is the vertical line  . Since the parabola is symmetrical, the horizontal distance from the vertex back to the point where the projectile started, namely ) , is the same as the horizontal distance from the vertex to the point where the ball hits the ground again (and presumably, for this rather simplistic model anyway, the ball stops). Since the ball started at a horiziontal distance of 0 and was at the vertex when it reached a horizontal distance of 25, it must have continued for another 25 feet before hitting the ground (0 height) and coming to rest. Hence, the maximum horizontal travel was 25 + 25 = 50 feet.
You might care to notice that if you were to set your function equal to zero and solve the quadratic, you would have two roots, namely 0 and 50.
John
My calculator said it, I believe it, that settles it
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