A tank is filled with water up to a height \(\mathrm{H}\). Water is allowed to come out of a hole P in one of the walls at a depth \(\mathrm{D}\) below the surface of water express the horizontal distance \(\mathrm{x}\) in terms of \(\mathrm{H}\) and \(\mathrm{D}\). (B) $\left.\mathrm{x}={ }^{\alpha} \sqrt{[}\\{\mathrm{D}(\mathrm{H}-\mathrm{D})\\} / 2\right]$ (D) \(\mathrm{x}=4[\mathrm{D}(\mathrm{H}-\mathrm{D})]\)

Short Answer

Expert verified
The short answer is: \(x = \sqrt{4D(H - D)}\).

Step by step solution

01

Find the Initial Vertical Velocity

To calculate the initial velocity of the water as it leaves the hole, we can use Torricelli's theorem, which states that the speed of the fluid at the hole is equal to the speed it would acquire if it fell freely from a height equal to the depth of the hole. Therefore, the velocity of the water, v, can be found using the equation: \(v = \sqrt{2gD}\), where g is the acceleration due to gravity (approximately 9.81 m/s^2).
02

Calculate Time to Hit the Ground

As the water leaves the hole horizontally with velocity v, it will also be subject to gravity, which will make it fall towards the ground. We can use kinematic equations to find the time it takes for the water to reach the ground. In this case, we have the following equation: \((H - D) = \frac{1}{2}gt^2\), where t is the time it takes for the water to hit the ground. We need to solve this equation for t: \(t = \sqrt{\frac{2(H - D)}{g}}\).
03

Calculate the Horizontal Distance x

Now that we have the initial horizontal velocity of the water and the time it takes to reach the ground, we can calculate the horizontal distance x: \(x = vt\), where x is the horizontal distance, v is the horizontal velocity, and t is the time found in step 2. Substitute the values for v and t from steps 1 and 2: \(x = \sqrt{2gD} \cdot \sqrt{\frac{2(H - D)}{g}}\), \(x = \sqrt{2gD \cdot \frac{2(H - D)}{g}}\). Simplifying the expression, we get: \(x = \sqrt{4D(H - D)}\). The correct answer is (D) \(x=4[D(H-D)]\).

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