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January 25, 2015

January 25, 2015

Posted by **Candy** on Thursday, January 29, 2009 at 9:33pm.

- Calculus -
**bobpursley**, Thursday, January 29, 2009 at 9:44pmI am not certain of your second term, and it affects my thinking. The .x^2, confuses me.

But nevertheless, if x is the horizontal axis, I would take the derivative of the equation as dx/dm when y is zero (striking the axis)

mx= second term

take the derivative with respect to m and solve for x (Max x)

Then, put that value of x into the original equation, with y zero, and solve for m.

- Calculus -
**Reiny**, Thursday, January 29, 2009 at 9:52pmI am going to read your equation as

y = mx - (e^(2m)/1000)x^2

first let's find where the ball will strike the horizontal axis, that is, at y=0

so mx - (e^(2m)/1000)x^2 = 0

x(m - (e^(2m)/1000)x) = 0

so x = 0 (at the origin, we knew that)

and

x = 1000m/(e^(2m))

so the distance from the origin is

d = 1000m/(e^(2m))

so dd/dm = [e^2m(1000) - 1000m(2)(e^2m)]/(e^2m)^2

= 0 for a max of d

we get

e^2m(1000 - 2000m) = 0

e^2m = 0 has no solution OR

1000 = 2000m

m = 1/2

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