calculus Question, please help me

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<p>actually i tried to simplify the question down, here is the actual question</p>

<p>A ball is thrown from the origin of a coordinate system. the equation of its path is y = mx - [(e^(2m))(x^2) / 1000 ], 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? Justify your answer.</p>

<p>i must have approached this problem wrong. how must i begin in solving this problem?</p>

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<p>I'd take the first derivative, find the point at which Y is the highest (the max), and then multiply that x by 2.</p>

<p>But isn't the slope in the end just going to be -m?</p>

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<p>Lets call the value of x where the ball strikes the horizontal axis s. The value of y at that point is 0 by definition. Substituting that into the equation of motion we get</p>

<p>ms = e^(2m)s^2/1000
since we are not considering the trivial case where s = 0, we can divide both sides by s and get</p>

<p>m = e^(2m)s/1000 or</p>

<p>s = 1000 m / e^(2m)</p>

<p>then it is a relatively simple matter of taking ds/dm and setting it to 0 to find the maximum value of s.</p>

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<p>but how do you find ds/dm??
ds/dm = 1000e^(2m) - 2000e^(2m) / e^4(m^2) (by quotient rule)
0=1000e^(2m) - 2000e^(2m) / e^4(m^2)
So, 1000e^(2m) = 2000e^(2m)
e^(2m)=2e^(2m)
0=2 ??? but i need a value for m
please do this problem and show your steps</p>

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<p>rewrite it as s = 1000m * e^-(2m) and do a product rule</p>

<p>(1000m) * e^-(2m) * -2 + (1000) * e^-(2m) = 0
(-2000m + 1000)/e^-(2m) = 0
-2000m + 1000 = 0
m = .5</p>

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<p>By quotient rule</p>

<p>ds/dm = 1000e^(2m) - 2000m * e^(2m) / e^(4m)
0 = 1000e^(2m) - 2000m * e^(2m) / e^(4m)
0 = 1000e^(2m) - 2000m * e^(2m)
factor out a e^(2m) and divides both sides by it
0 = 1000 - 2000m
m = .5</p>

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<p>thanks a lot BCC, i should have factored out</p>