Question

There are two holes $O_1$ and $O_2$ in a tank of height $h_3$. Water emerging from $O_1$ and $O_2$ strikes the ground at the same points as shown in figure, then

• A. $H=h_1+h_2$
• B. $H=h_2-h_1$
• C. $H=\sqrt{h_1{h}_2}$
• D. None of these

Answer 1

The correct option is A $H=h_1+h_2$

Velocity of water from hole at any depth $\left(h\right)$ from the free surface of water is $\sqrt{2gh}$ which is intially in $x$ - drection.
Range of water becomes,
$R=velocity\times time$ ($\because$ acceleration in $x$-direction is zero)
We can calculate time by using third equation of motion in vertical direction ($y-direction$),
$s_y=u_{y}t+\frac{1}{2}a{t}^2$ (assuming physical quantitiy to be positive in downward direction)
Initial velocity in $y$- direction, $u_y=0\frac{m}{s}$
Distance travel in $y$-direction, $s_y=h_3-h$
$⟹h_3-h=\frac{1}{2}g{t}^2$
$t=\sqrt{\frac{2(h_3-h)}{g}}$
Hence Range (R) becomes,
$R=u_{x}t=\sqrt{2gh}\times \sqrt{\frac{2(h_3-h)}{g}}=2\sqrt{h(h_3-h)}$
$\because$ range is equal for both jet,
$⟹R_1=R_2$
$2\sqrt{h_1(h_3-h_1)}=2\sqrt{h_2(h_3-h_2)}$
$h_1(h_3-h_1)=h_2(h_3-h_2)$
$h_3{h}_1-h_1^2=h_3{h}_2-h_2^2$
$h_1^2-h_2^2=h_3{h}_1-h_3{h}_2$
$(h_1+h_2)(h_1-h_2)=h_3(h_1-h_2)$
$h_3=h_1+h_2$