Chapter 10, Problem 49P

(a)

To determine

To Calculate: The horizontal distance from the base of the tank will the fluid strike the ground.

Answer to Problem 49P

The horizontal distance from the base of the tank when fluid strikes the ground is $\sqrt{2g\left(h_2-h_1\right)}\sqrt{\frac{2h_1}{g}}$ and the height $h_1^{\text{'}}$ where the hole is placed is $\left(h_2-h_1\right)$ .

Explanation of Solution

Given:

The height of the opening of the tank is $h_1$ and the height of the liquid surface is $h_2$ . Initially, the tank is at the rest.

Formula used:

The continuity equation is,

  $Av=\text{const}\text{.}$

Here, $A$ is the area of the cross-section of the hole,

  $v$ is the velocity of the fluid.

The Bernoulli’s equation is,

  $P_1+\frac{1}{2}\rho v_1{}^2+\rho g{h}_1=P_2+\frac{1}{2}\rho v_2{}^2+\rho g{h}_2$

Here, $P_1$ is the pressure of fluid in the tank at point 1,

  $P_2$ is the pressure of fluid in the tank at point 2,

  $\rho$ is the density of the fluid,

  $h_1$ is the height from the reference to point 1,

  $h_2$ is the height from the reference to point 2,

  $v_1$ is the velocity of liquid at point 1,

  $v_2$ is the velocity of liquid at point 2,

As pressure of liquid is same at both the points so $P_1=P_2$ ,

And at speed at point 2 is zero. Thus, $v_2=0$ .

On substituting the values in the above expression,

  $\begin{array}{l}\frac{v_1{}^2}{2}=g\left(h_2-h_1\right)\\ v_1=\sqrt{2g\left(h_2-h_1\right)}\end{array}$

Calculation:

When the liquid is launched horizontally, the initial speed $v_1$ is zero. By using the Bernoulli’s equation for the constant acceleration to find the time of fall with upward as the positive direction, then the horizontal speed is

  $\begin{array}{l}y=y_1+v_{1}t+\frac{1}{2}g{t}^2\\ 0=h_1+0-\frac{1}{2}g{t}^2\\ t=\sqrt{\frac{2h_1}{g}}\end{array}$

Write the expression for the change in the height.

  $\Delta h=v_{1}t$

Substitute the values of $v_1$ and $t$ in the above expression.

  $\begin{array}{c}\Delta t=\sqrt{2g\left(h_2-h_1\right)}\sqrt{\frac{2h_1}{g}}\\ =2\sqrt{\left(h_2-h_1\right)h_1}\end{array}$

Conclusion:

The horizontal distance from the base of the tank is $\Delta h=2\sqrt{\left(h_2-h_1\right)h_1}$ .

(b)

To determine

To Calculate: The other height when a hole can be placed.

Answer to Problem 49P

The other height is $h_1^{\text{'}}=h_2-h_1$ , when a hole can be placed so that the emerging liquid will have the same range.

Explanation of Solution

Given:

The height of the opening of the tank is $h_1$ and the height of the liquid surface is $h_2$ . Initially, the tank is at the rest. Assume, $v_2\approx 0$

Calculation:

The horizontal distance from the base of the tank is $\Delta h=2\sqrt{\left(h_2-h_1\right)h_1}$ .Here, the height $h_1^{\text{'}}$ is calculated:

  $\begin{array}{l}2\sqrt{\left(h_2-h_1\right)h_1}=2\sqrt{\left(h_2-h^{\text{'}}{}_1\right)h^{\text{'}}{}_1}\Rightarrow \left(h_2-h_1\right)h_1=\left(h_2-h^{\text{'}}{}_1\right)h^{\text{'}}{}_1\\ \Rightarrow h_1^{\text{'}}{}^2-h_2{h}_1^{\text{'}}+\left(h_2-h_1\right)h_1=0\\ \Rightarrow h_1^{\text{'}}=\frac{h_2\pm \sqrt{h_2^2-4\left(h_2-h_1\right)h_1}}{2}\\ =\frac{h_2\pm \sqrt{h_2^2-4h_1{h}_2+4h_1{}^2}}{2}=\frac{h_2\pm \left(h_2-2h_1\right)}{2}\\ =\frac{2h_2-2h_1}{2},\frac{2h_1}{2}\end{array}$

Hence, the height is,

  $h_1^{\text{'}}=h_2-h_1$

Conclusion:

The height $h_1^{\text{'}}$ is $h_2-h_1$ .