Water stands at a depth H in a large, open tank whose side walls are vertical (Fig. P12.77). A hole is made in one of the walls at a depth h below the water surface. (a) At what distance R from the foot of the wall does the emerging stream strike the floor? (b) How far above the bottom of the tank could a second hole be cut so that the stream emerging from it could have the same range as for the first hole?

The given data can be listed below as,

• The height of hole is,$\left(H-h\right)$.
• The water depth is,$H$.
• The depth of hole from top surface is, h.
• The distance of emerging stream strike the floor is, R.

The second equation of motion establishes a relation between distance travelled, initial velocity, time taken, and acceleration.

$\mathbf{\alpha }\mathbf{=}\mathbf{ut}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{gt}}^{\mathbf{2}}$

The water is in free fall, after separation from the tank, the acceleration in x-direction is zero and y-direction is g. Then the speed of efflux is express as,

$V=\sqrt{2gh}$

The initial velocity of water is zero, then time the takenby the water to reach the ground can becalculated as,

$$\begin{gathered} d=ut+\frac{1}{2}g{t}^2 \\ \left(H-h\right)=0\times t+\frac{1}{2}g{t}^2 \\ t=\sqrt{\frac{2\left(H-h\right)}{g}} \end{gathered}$$

The distance travelled by water is expressed as,

$R=vt$

Substitute all the value in the above equation.

$$\begin{gathered} R=\sqrt{2gh}\times \sqrt{\frac{2\left(H-h\right)}{g}} \\ =\sqrt[2]{h\left(H-h\right)} \end{gathered}$$

Hence the water travels horizontal distance is$\sqrt[2]{h\left(H-h\right)}$.

If depth of hole from the top of the surface is h’, then the speed of efflux is express as,

$V=\sqrt{2{\mathrm{h}}^{\text{'}}}$

The initial velocity of the water is zero, then time taken by the water to reach the ground will be,

$$\begin{gathered} d=ut+\frac{1}{2}g{t}^2 \\ \left(H-h\right)=0\times t+\frac{1}{2}g{t}^2 \\ t=\sqrt{\frac{2\left(H-h\right)}{g}} \end{gathered}$$

The water travels horizontal distance is express as,

$$\begin{gathered} R^{\text{'}}=vt \\ {R}^{\text{'}}=\sqrt{2g{h}^{\text{'}}}\times \sqrt{\frac{2\left(H-h^{\text{'}}\right)}{g}} \\ {R}^{\text{'}}=2\sqrt{h^{\text{'}}\left(H-h^{\text{'}}\right)} \end{gathered}$$

Since the range of the water remains same for both the holes, therefore we can write,

$$\begin{gathered} R=R^{\text{'}} \\ \sqrt[2]{h\left(H-h\right)}=2\sqrt{h^{\text{'}}\left(H-h^{\text{'}}\right)} \\ h\left(H-h\right)=h^{\text{'}}\left(H-h^{\text{'}}\right) \\ {h}^{\text{'}}=H-h \end{gathered}$$

Thus, the height of the second hole should be $H-h$.