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- Thread starter MarkFL
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- Feb 14, 2012

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Hi **MarkFL**,

I'm hoping to get some hint on how to at least start to work on the problem...

I'm hoping to get some hint on how to at least start to work on the problem...

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Well, since you asked so nicely...sure no problem!HiMarkFL,

I'm hoping to get some hint on how to at least start to work on the problem...

\(\displaystyle \tag{1}x=v_0\cos(\theta)t\)

\(\displaystyle \tag{2}y=-\frac{g}{2}t^2+v_0\sin(\theta)t\)

To eliminate the parameter $t$, we solve (1) for $t$ and substitute into (2) to get:

\(\displaystyle \tag{3}y=\tan(\theta)x-\frac{g}{2v_0^2\cos^2(\theta)}x^2\)

Using \(\displaystyle \frac{1}{ \cos^2(\theta)}= \sec^2(\theta)= \tan^2(\theta)+1\) (3) becomes:

\(\displaystyle \tag{4}y=\tan(\theta)x-\frac{g}{2v_0^2}\left(\tan^2(\theta)+1 \right)x^2\)

Now, using the substitution:

\(\displaystyle u=\tan(\theta)\)

we get the parametrization:

\(\displaystyle \tag{5}y=ux-\frac{g}{2v_0^2}\left(u^2+1 \right)x^2\)

Now, let the boundary we seek be denoted by \(\displaystyle f(x)\). What can we say about the intersection of $y$ and $f(x)$?

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\(\displaystyle u^2-\frac{2v_0^2}{gx}u+\left(1+\frac{2v_0^2f(x)}{gx^2} \right)=0\)

Because $y$ will only ever be tangent to $f(x)$ as it is the boundary, we know therefore that the discriminant must be zero, giving us:

\(\displaystyle \left(-\frac{2v_0^2}{gx} \right)^2-4\left(1+\frac{2v_0^2f(x)}{gx^2} \right)=0\)

\(\displaystyle \left(\frac{v_0^2}{gx} \right)^2=1+\frac{2v_0^2f(x)}{gx^2}\)

And then solving for $f(x)$, we find:

\(\displaystyle f(x)=-\frac{g}{2v_0^2}x^2+\frac{v_0^2}{2g}\)