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Step-by-Step Solution
Step 1: Identify the physical situation
The tank is filled with water up to a height of 12 m. A small hole is made in the wall at a depth h below the water surface. We need to find the value of h for which the horizontal range (the distance on the ground where the water lands) is maximized.
Step 2: Express the exit velocity of water
From Torricelli's theorem (or by applying Bernoulli’s principle for an efflux out of a small orifice), the speed v of water emerging from the hole is given by:
v = \sqrt{2 g h}
where g is the acceleration due to gravity and h is the depth of the hole below the free surface of water.
Step 3: Determine the time of flight
The hole is at depth h from the top, so the vertical distance of the hole above the ground is (12 - h) (since the total height of the water column is 12 m). The time t it takes for the water stream to fall from height (12 - h) to the ground under gravity is given by the free-fall equation:
t = \sqrt{\frac{2(12 - h)}{g}}
Step 4: Write the expression for the horizontal range
The horizontal range R is the product of horizontal velocity and the time of flight. Since the stream exits horizontally from the hole, the horizontal velocity is simply v = \sqrt{2gh} . Thus,
R = \bigl(\sqrt{2gh}\bigr) \times \sqrt{\frac{2(12 - h)}{g}}
On simplifying,
R = 2 \sqrt{h (12 - h)}
Step 5: Maximize the function for range
To find the maximum value of R , we note that R depends on h through the product h(12 - h) . We can analyze this using the idea that a product h(12 - h) is maximized when h = 12 - h , i.e. h = 6 . Formally, we can write:
f(h) = h(12 - h)
f(h) = 12h - h^2
\frac{df}{dh} = 12 - 2h = 0 \;\Longrightarrow\; h = 6
This is the point at which h(12 - h) is maximum.
Step 6: Conclude the optimal depth
Therefore, the depth h at which the emerging water stream hits the ground at the maximum distance is 6\,\text{m} below the water’s surface.
