Question 77574
As he plugged the ball toward left field, it rose from 3 to 10 to 50 feet as the horizontal distance from home plate increased from 0 to 2 to 14 feet. Assume that the vertical distance varies quadratically with the horizontal distance.
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You have three tracking points: (0,3),(2,10),(14,50)
The equation is y=ax^2+bx+c
Substitute the point values and solve for a,b,c
3=0a+0b+c
10=4a+2b+c
50=196a+14b+c
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Solve the system of linear equations.
I used a TI calculator's matrix method to get:
X=A^-1B
a= -0.0119047
b= 3.5238095
c= 3
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EQUATION:
y=-0.119047x^2+3.5238095x+3
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Cheers,
Stan H.