A tank is filled by water (`rho = 10^3 kg//m^3`). A small hole is made in the side wall of the tank at depth 10 m below water surface. A water jet emerges horizontal from the hole and falls at horizontal distance R from it. The amount of extra pressure (in terms of atmospheric pressure) that must be applied on the water surface, so that range becomes 3R on the on the ground will be (cross section area of hole is negligible and 1 atm = 105 Pa, g = 10 m/`s^2`)

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the initial velocity of the water jet (V1)
The velocity of the water jet emerging from the hole can be calculated using Torricelli's theorem, which states that the speed (V) of fluid flowing out of an orifice under the influence of gravity is given by:

\[ V_1 = \sqrt{2gh} \]
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