# Where can the water jet spray?

#### MarkFL

Staff member
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

##### MHB POTW Director
Staff member
Hi MarkFL,

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

#### MarkFL

Staff member
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!

Hint:

I would begin with the parametric equations of motion:

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

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

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 $f(x)$?

#### MarkFL

Staff member
Solution:

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

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

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:

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