The problem is about fluid dynamics and the trajectory of a jet of water. The maximum horizontal distance (range) a jet of water can reach is when it is projected at an angle of 45 degrees.

1. First, we need to find the velocity of the water as it leaves the orifice. This can be found using Torricelli's theorem, which states that the speed v of efflux of a fluid under the force of gravity from a hole in a large, open container is given by v = sqrt(2gh), where g is the acceleration due to gravity and h is the height of the fluid above the hole.

2. Next, we need to find the angle at which the water leaves the tank. This is given by the equation tan(theta) = h / (h - h'), where h' is the height of the hole above the ground.

3. We set this angle to 45 degrees for maximum range, so tan(45) = 1 = h / (h - h'). Solving for h' gives h' = h / 2.

4. Therefore, the hole should be placed at a height h' = h / 2 = 2.8m / 2 = 1.4m above the ground for the jet to strike the ground at a maximum distance from the tank.

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The problem is about fluid dynamics and the trajectory of a jet of water. The maximum horizontal distance (range) a jet of water can reach is when it is projected at an angle of 45 degrees.

1. First, we need to find the velocity of the water as it leaves the orifice. This can be found using Torricelli's theorem, which states that the speed v of efflux of a fluid under the force of gravity from a hole in a large, open container is given by v = sqrt(2gh), where g is the acceleration due to gravity and h is the height of the fluid above the hole.

2. Next, we need to find the angle at which the water leaves the tank. This is given by the equation tan(theta) = h / (h - h'), where h' is the height of the hole above the ground.

3. We set this angle to 45 degrees for maximum range, so tan(45) = 1 = h / (h - h'). Solving for h' gives h' = h / 2.

4. Therefore, the hole should be placed at a height h' = h / 2 = 2.8m / 2 = 1.4m above the ground for the jet to strike the ground at a maximum distance from the tank.

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The problem is about fluid dynamics and the trajectory of a jet of water. The maximum horizontal distance (range) a jet of water can reach is when it is projected at an angle of 45 degrees.

1. First, we need to find the velocity of the water as it leaves the orifice. This can be found using Torricelli's theorem, which states that the speed v of efflux of a fluid under the force of gravity from a hole in a large, open container is given by v = sqrt(2gh), where g is the acceleration due to gravity and h is the height of the fluid above the hole.

2. Next, we need to find the angle at which the water leaves the tank. This is given by the equation tan(theta) = h / (h - h'), where h' is the height of the hole above the ground.

3. We set this angle to 45 degrees for maximum range, so tan(45) = 1 = h / (h - h'). Solving for h' gives h' = h / 2.

4. Therefore, the hole should be placed at a height h' = h / 2 = 2.8m / 2 = 1.4m above the ground for the jet to strike the ground at a maximum distance from the tank.

A tank 2.8 m high standing on the ground is kept full of.water. There is an orifice in itsvertical side at a depth hm below the surface of water, Find the vallue of h in order that thejet may strike the ground at a maximum distance from the tank

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