Question

A water jet is being used to fill a cylindrical container of diameter $1m$ and height $2m$. The jet shoots the water at an angle of ${45}^o$ with velocity $u$ above the same horizontal plane on which the container is kept. The distance between the jet and container is $6m$ as shown in the diagram. For which of the following value(s) of $u$, the water will fill up the container. (Take $g=10m/s^2)$

• A. $\sqrt{92}$
• B. $\sqrt{67}$
• C. $\sqrt{97}$
• D. $\sqrt{62}$

Answer 1

Answer: (A), (C)

$\sqrt{97}$


Let the nozzle of the water jet be at the origin. Then the nearest part of the rim of the tank is at $(x,y)\equiv (6,2)$ and the furthest part of the rim is at $(x,y)\equiv (7,2)$. The trajectory of the water can be found as follows:

$x\left(t\right)=\frac{\sqrt{2}}{2}{v}_{0}t\left(\cos \left({45}^o\right)=\frac{\sqrt{2}}{2}\right)$$y\left(t\right)=\frac{\sqrt{2}}{2}{v}_{0}t-\frac{1}{2}g{t}^2\left(\sin \left({45}^o\right)=\frac{\sqrt{2}}{2}\right)$
Substituting the $\left(t=\frac{\sqrt{2}x}{v_{\mathrm{0\; from\; first\; equation}}}\right)$ into the second equation gives the trajectory.

$y=x-\left(\frac{g}{{v_0}^2}\right)x^2$
Now we want $y$ to be equal to $2$ when $x$ is between $6$ and $7$ but also that we are on the downward path of the parabola there.

Now
$x-\left(\frac{g}{{v_0}^2}\right)x^2=2$can be written as $\left(\frac{g}{{v_0}^2}\right)x^2-x+2=0$
with solutions:
$x=\frac{{v_0}^2}{2g}\pm \frac{{v_0}^2}{2g}\sqrt{\left(1-\frac{8g}{{v_0}^2}\right)}$but only the positive sign solution corresponds to the downward part of the parabola.
so we know that
$6<\frac{{v_0}^2}{2g}+\frac{{v_0}^2}{2g}\sqrt{\left(1-\frac{8g}{{v_0}^2}\right)}<7$
solving ${\left(12g-{v_0}^2\right)}^2<{v_0}^2-8g\sqrt{\left(1-\frac{8g}{{v_0}^2}\right)}<{(14g-{v_0}^2)}^2$
Left inequality : ${v_0}^4-24g{v_0}^2+{\left(12g\right)}^2<{v_0}^4-8g{v_0}^2{v}_0>\sqrt{9g}$Right inequality:
${v_0}^4-8g{v_0}^2<{v_0}^4+{\left(14g\right)}^2-28g{v_0}^2{v}_0<\sqrt{\frac{49}{5}g}$so the range of the initial projected velocity will be
$\sqrt{9g}<v_0<\sqrt{\frac{49}{5}g}$

For detailed solution watch the next video.