# Finding the range of launch speeds for which water shot from a hose will enter a tank a distance away

A water hose is used to fill a large cylindrical storage tank of diameter $$D$$ and height $$2D$$. The hose shoots the water at $$45^\circ$$ above the horizontal from the same level as the base of the tank and is a distance $$6D$$ away. For what range of launch speeds $$v_0$$ will the water enter the tank. Ignore air resistance.

I've made a graphic representing the problem (with $$D=1$$) showing the trajectory of the water at the minimum launch speed $$v_{0\text-m}$$ (blue) and maximum launch speed $$v_{0\text-M}$$ (red) as well as the tank (light gray):

Clearly I've made a mistake in solving for the minimum launch speed - I expect the blue parabola to intersect the rectangle at its top left vertex. I'm not sure where the mistake lies, because as far as I can tell I've used the same approach to solve for $$v_{0\text-m}$$ as I had for $$v_{0\text-M}$$, which I'll show first.

The position vector for water particles escaping the hose has components

$$\begin{cases} x=v_{0\text-M}\cos45^\circ t\\ y=v_{0\text-M}\sin45^\circ t-\frac g2t^2 \end{cases}$$

With $$x=7D$$, solve for $$t$$ in the first equation, substitute into the second with $$y=2D$$ to get

$$t=\frac{7D}{v_{0\text-M}\cos45^\circ}\implies2D=v_{0\text-M}\sin45^\circ\left(\frac{7D}{v_{0\text-M}\cos45^\circ}\right)-\frac g2\left(\frac{7D}{v_{0\text-M}\cos45^\circ}\right)^2$$

$$\implies5=\frac{49Dg}{v_{0\text-M}^2}\implies v_{0\text-M}=7\sqrt{\frac{Dg}5}$$

Now to find $$v_{0\text-m}$$, the only change in the work above - I would think - would be to set $$x=6D$$.

$$t=\frac{6D}{v_{0\text-m}\cos45^\circ}\implies2D=v_{0\text-m}\sin45^\circ\left(\frac{6D}{v_{0\text-m}\cos45^\circ}\right)-\frac g2\left(\frac{6D}{v_{0\text-m}\cos45^\circ}\right)^2$$

$$\implies2=\frac{12Dg}{v_{0\text-m}^2}\implies v_{0\text-m}=\sqrt{6Dg}$$

Manipulating the code for the plot of the blue parabola, it would appear the correct answer for $$v_{0\text-m}$$ is closer to $$3\sqrt{Dg}$$. How can I salvage my attempt to obtain the right solution?

• – amd Oct 12 '18 at 1:22
• The blue trajectory is incorrect. It should reach the top left of the water tank! – hypergeometric Jan 21 '19 at 16:22
• @hypergeometric Yes, as I mentioned, "I expect the blue parabola to intersect the rectangle at its top left vertex". I found the source of the discrepancy already; see my answer below. – user170231 Jan 21 '19 at 17:23

Minor algebraic mistake...

$$2D=v_{0\text-m}\sin45^\circ\left(\frac{6D}{v_{0\text-m}\cos45^\circ}\right)-\frac g2\left(\frac{6D}{v_{0\text-m}\cos45^\circ}\right)^2$$

should reduce to

$$2D=6D-\frac g2\frac{36D^2}{v_{0\text-m}^2\cos^245^\circ}\implies4D=\frac{36D^2g}{v_{0\text-m}^2}\implies v_{0\text-m}=3\sqrt{Dg}$$

At some point I decided to divide through all terms by $$2D$$ except for one, which I had mistakenly divided by $$3D$$ instead.

The equation of trajectory of water jet is $$y = x-\dfrac{g}{v^2} x^2$$. Let $$x$$ and $$y$$ be coordinates of any point in water jet at a height of $$2D$$. So water to fall we must have $$6D < x < 7D$$. By putting $$y = 2D$$ in above equation we get quadratic in $$x$$ as $$\dfrac{g}{v^2} x^2 - x +2D = O$$. Discard negative value of $$x$$ take only positive value then put value of $$x$$ in above inequality and solve you will get

$$\sqrt{9gD} < v < \sqrt{9.8gD}$$