Leaky Water Tank

In this problem, we have to use Bernoulli's Equation and the Equation of Continuity .

Bernoulli's Equation: $p_1+\rho gh_1+\frac{1}{2}\rho{v_1}^2=p_2+\rho gh_2+\frac{1}{2}\rho{v_2}^2$

Equation of Continuity: $S_1 v_1=S_2 v_2$ where $Sv$ is the Flow Rate .

$\text{LHS}$ is the case for the tank while $\text{RHS}$ is the case for the hole in the tank.

In this problem, we have to make some assumptions:

1. As the surface area of the tank, $S_1$ is very large compare to the surface area of the hole $S_2$ , we assumed that $v_1\approx0$

2. The pressure, $p_1$ and $p_2$ are the same as they both are in the same environment.

3. We take $h_2$ as $0$ for easier calculation. Relatively, $h_1$ will be $10cm$

4. The density of the liquid inside the tank, $\rho$ is the same.

By these assumption, we can get a simpler equation:

$gh_1=\frac{1}{2}{v_2}^2$

Substitute the given information we can get

$v=14\text{ cms}^{-1}$

The direction of the water coming out from the hole is horizontal, thus

$s_y=\frac{1}{2}gt$

$t=2.474\text{ s}$

Solve for $x$ ,

$x=vt$

$x=34.64\text{ cm}$

$M(drop)gh\quad =\quad \frac{ M{ (drop)v }^{ 2 } }{ 2 } \quad thus\quad v\quad =\quad \sqrt [ 2 ]{ 2gh } \\ to\quad calculate\quad v,\quad lets\quad use\quad cinematics:\\ -\quad in\quad the\quad x\quad axis\quad we\quad have:\\ \quad x=\quad v\quad \times \quad t\quad \rightarrow \quad x\quad =\quad \sqrt [ 2 ]{ 2gh } \times \quad t\quad (1)\\ -\quad in\quad the\quad y\quad axis:\\ y=\quad g\frac { { t }^{ 2 } }{ 2 } \quad =\quad H-h\quad thus\quad t\quad =\sqrt { \frac { 2(H-h) }{ g } } \quad (2)\\ Substitute\quad (2)\quad in\quad (1):\\ x=\quad \sqrt [ 2 ]{ 2gh } \times \sqrt { \frac { 2(H-h) }{ g } } =\quad 2\times \sqrt { (H-h)\times h } \\ =\quad 2\times \sqrt { (40-10)\times 10 } =34.641\quad cm$

For a water particle coming out with a velocity of u cm/sec, we may write: mgh = (1/2) mu² or u² = 2gh =20g since h=10 cm General eqn. of parabolic motion: y =xtan(θ) – gx²/(2u²cos²(θ)) where y = -30, θ = 0 so that – 30 = x²/(2*20) or x² =1200 or x~34.641 cm