Question

A tank is filled with water to a height h. Two holes are made on its side wall, one at a height of h from the bottom and the other at a depth h from the top. The horizontal jets starting from the two holes meet the ground outside (in level with the bottom of the tank) at the same point. The distance of this point from the side of the tank is-

Answer 1

If both jets falling at same point than both is also equal to height of the hole h...

Answer 2

A tank is filled with water to a height h. Two holes are made on its side wall, one at a height of h from the bottom and the other at a depth h from the top. The horizontal jets starting from the two holes meet the ground outside (in level with the bottom of the tank) at the same point. The distance of this point from the side of the tank is

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

Explanation:

From Bernoulli's Theorem we know that if a jet is coming out of a hole on the site of a tank and if this hole is at distance h from the top of the liquid surface then the velocity of the jet is

$v=\sqrt{2gh}$

Therefore, the velocity of the top jet will be

$v_1=\sqrt{2gh}$

And, the velocity of the bottom jet will be

$v_2=\sqrt{2g(H-h)}$

The first jet will travel the height H-h before it reaches the ground and the second jet travels height h before it reached the ground

If the time taken by the first jet is t then

Since there is no vertical component of the velocity for the jet

Therefore, by using the second equation of motion

$H-h=0\times t+\frac{1}{2}gt^2$

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

The jet will come out of the hole horizontally and since there is no acceleration in the horizontal direction

The horizontal distance travelled by the jet is given by

Distance = Velocity x Time

or, $d=\sqrt{2gh}\times \sqrt{\frac{2(H-h)}{g}$

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

Hope this answer is helpful.

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