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:

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

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

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]

Solution:

Spoiler

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:

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

 

[CellCoree]

The water jet can reach a maximum height when it is sprayed at a 45 degree angle ($\theta = \pi/4$) from the horizontal axis. This is because at this angle, the horizontal velocity ($v_x$) and vertical velocity ($v_y$) are equal, resulting in the maximum range of the water jet.

Using the kinematic equations, we can find an equation that describes the boundary between where the water can reach and where it cannot. Let's assume that the water jet is sprayed from a distance of $d$ from the base point and the initial velocity of the water is $v_0$.

At any given time $t$, the horizontal position $x$ of the water jet can be described by $x = d + v_0 \cos \theta \, t$. Similarly, the vertical position $y$ can be described by $y = v_0 \sin \theta \, t - \frac{1}{2} g t^2$, where $g$ is the acceleration due to gravity.

To find the boundary where the water can reach, we need to find the time $t$ at which the water hits the ground, i.e. when $y=0$. Solving for $t$ in the second equation, we get $t = \frac{2v_0 \sin \theta}{g}$. Substituting this value of $t$ in the first equation, we get the equation for the boundary as:

$x = d + \frac{2v_0^2 \cos \theta \sin \theta}{g}$

This equation represents the maximum horizontal distance $x$ that the water can reach from the base point $d$ at any given angle $\theta$. Any point beyond this distance will be out of reach for the water jet.

In conclusion, the equation for the boundary between where the water can reach and cannot reach is $x = d + \frac{2v_0^2 \cos \theta \sin \theta}{g}$.