MHB Where can the water jet spray?

[MarkFL]

Consider a water jet that sprays from ground level on flat ground. The jet can spray at any angle $0<\theta<\pi$ along some horizontal axis, that is in one vertical plane. The jet sprays water at a velocity of $v_0$. Ignoring drag, find an equation that describes the boundary between where water can reach and where it cannot.

 

[anemone]

Hi MarkFL,

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

 

[MarkFL]

Hi MarkFL,

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

Well, since you asked so nicely...sure no problem! (Sun)

Hint:

Spoiler

I would begin with the parametric equations of motion:

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

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

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

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

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

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

Now, using the substitution:

$$u=\tan(\theta)$$

we get the parametrization:

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

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

 

[MarkFL]

Solution:

Spoiler

Continuing from the hint I gave above, we will equate $y=f(x)$ and obtain the quadratic in $u$ in standard form:

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

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:

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

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

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

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