Calculus
posted by Candy on .
A ball is thrown from the origin of a coordinate system. The equation of its path is y = mx  e^2m/1000 . x^2, where m is positive and represents the slope of the path of the ball at the origin. For what value of m will the ball strike the horizontal axis at the greatest distance from the origin?

I 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. 
I 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