Question

A rectangular tank of height $6m$ filled with water is placed near the bottom of an incline of angle ${30}^{\circ }$. At height $x$ from bottom a small hole is made (as shown in figure ) such that stream coming out from hole strikes the inclined plane normally. Find $x$ in meter .

Answer 1

The velocity of stream when it gets out of the hole is $V_o=\sqrt{2g(6-x)}$

Now lets say that the stream strikes the inclined plane after t seconds,

the horizontal component of the velocity will still be, $V_x=V_0$

Now, the angle of incline is ${30}^o$, so the stream will be forming an angle of $90-30={60}^o$ on striking the inclined plane.

Now, $V_y/V_x=\tan {60}^o=\sqrt{3}$

Therefore, $V_y=\sqrt{3}{V}_x=\sqrt{3}{V}_0$

Since, the initial velocity in the vertical direction is $0$,

$V_y=0+gt$

$gt=\sqrt{3}{V}_0$

Therefore, t = $\sqrt{3}{V}_0/g$

Now, the distance travelled in the vertical direction will be, $x-0.5g{t}^2$

Or, distance = $x-0.5g\times 3V_0^2/g^2$

Distance travelled in the horizontal direction = $V_{0}t$

The two distances are related as, $y/x=\tan {30}^o$

$x-0.5g{t}^2=\sqrt{1/3}{V}_{0}t$

Solving, this, $x=5h/6=5m$