MHB Where can the water jet spray?

  • Thread starter Thread starter MarkFL
  • Start date Start date
  • Tags Tags
    Jet Spray Water
AI Thread Summary
The discussion focuses on a water jet spraying from ground level at various angles in a vertical plane. The key problem is to derive an equation that defines the boundary of the spray reach based on the jet's initial velocity, v0, and the angle of spray, θ. Participants are seeking guidance on how to approach the problem mathematically. A hint is provided to assist in starting the solution process. The conversation emphasizes understanding the physics behind the water jet's trajectory.
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
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.
 
Mathematics news on Phys.org
Hi MarkFL,

I'm hoping to get some hint on how to at least start to work on the problem...:o
 
anemone said:
Hi MarkFL,

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

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

Hint:

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 $f(x)$?
 
Solution:

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:

$$f(x)=-\frac{g}{2v_0^2}x^2+\frac{v_0^2}{2g}$$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top