Water stands at a depth H in a large, open tank whose sidewalls are vertical. A hole is made in one of the walls at a depth h below the water surface.

Solution

Consider section-1 at the surface of the water and section-2 at the center of the hole.

Write the Bernoulli’s equation for the water inside the tank and for the water that flows out of it at a given depth.

$P_1+\frac{1}{2}\rho v_1^2+\rho gh=P_2+\frac{1}{2}\rho v_2^2+\rho g{h}_2$

Here, hh is depth of the hole from the water surface and $h_2$ is the height from datum where the water stream strikes, $\rho$ is the density of the water, and g is the acceleration due to gravity.

Substitute $P_{atm}$for $P_1$​ and $P_2$​, and 0 for $h_2$.

$P_{atm}+\frac{1}{2}\rho v_1^2+\rho gh=P_{atm}+\frac{1}{2}\rho v_2^2+\rho g(0)$

$\frac{1}{2}\rho v_1^2+\rho gh=\frac{1}{2}\rho v_2^2$

$\frac{1}{2}{v}_1^2+gh=\frac{1}{2}{v}_2^2$

Write the continuity equation.

$A_1{v}_1=A_2{v}_2$

Here, the area of the hole is significantly smaller than the area of the tank, so assume the speed at hole is negligible.

Further, solve for $v_2$

$v_2^2=(v_1^2+2gh)$

$v_2=\sqrt{v_1^2+2gh}$

Substitute 0 for$v_1$ and v for $v_2$

$v=\sqrt{2gh}v=2gh$

Since, in the given condition the water flows out and follows a trajectory under the influence of gravity. Thus, only gravitational acceleration is considered.

From kinematic equation of motion,

$(H-h)=\frac{g{t}^2}{2}$

Here, H is the height of the tank, h is the depth of the hole from the water surface, and t is time taken by the water stream to reach the ground.

Rearrange for t .

$t=\sqrt{\frac{2(H-h)}{g}}$

Write the expression for the horizontal distance travelled by the water.

R = vt

Here, v is the speed of the water and t is the time interval.

Substitute, $\sqrt{2gh}$​ for v and $\sqrt{\frac{2(H-h)}{g}}$ for t .

$R=\sqrt{2gh}(\sqrt{\frac{2(H-h)}{g}})$

$=\sqrt{\frac{2(H-h)2gh}{g}}$

$=2\sqrt{h(H-h)}$

(b)

Rewrite the expression for the range calculated in the step (1).

$R=[2\sqrt{h(H-h)}]$

Take square on both sides in the above equation.

$R^2=4h(H-h)$

$R^2=4Hh-4h^2$

Rearrange,

$4h^2+4Hh+R^2=0$

This is a quadratic equation with a variable h , thus, solve for the roots as,

$h=\frac{4H\pm \sqrt{16H^2-16R^2}}{8}$

$=\frac{1}{2}(H\pm \sqrt{H^2-R^2})$

Substitute H for R.

$h=\frac{1}{2}(H\pm \sqrt{H^2-H^2})$

$=\frac{H}{2}$