A tank of water 3 meters deep sits on a table 1 meter tall. A hole is located 2 meters from the bottom of the tank. Water flows out of the hole and strikes the ground. Where does it strike the ground? Assume incompressible, non viscous flow.
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To determine where the water strikes the ground, we need to consider the basic principles of fluid dynamics.
First, let's calculate the velocity of the water as it flows out of the hole. We can use Torricelli's law, which states that the velocity of an ideal fluid coming out of a small hole in a container is given by the equation v = √(2gh), where v is the velocity, g is the acceleration due to gravity (9.8 m/s²), and h is the height of the fluid above the hole.
In this case, the height of the fluid above the hole is 3 meters - 2 meters = 1 meter (the total height of the water minus the height from the hole to the ground). Plugging in the values, we find the velocity of the water leaving the hole is v = √(2 * 9.8 * 1) = √19.6 ≈ 4.43 m/s.
The water leaving the hole will follow a parabolic trajectory due to gravity. Since the water strikes the ground, its horizontal distance traveled (x) will depend on the time it takes to reach the ground.
To find the time it takes for the water to reach the ground, we can use the equation h = 0.5 * g * t², where h is the vertical distance (3 meters) and g is the acceleration due to gravity (9.8 m/s²). Solving for t, we get t = √(2h/g) = √(2 * 3 / 9.8) ≈ √0.6122 ≈ 0.78 seconds.
Now, we can calculate the horizontal distance (x) using the equation x = v * t, where v is the horizontal velocity (equal to the velocity of the water leaving the hole) and t is the time it takes to reach the ground. Plugging in the values, we find x ≈ 4.43 m/s * 0.78 s ≈ 3.45 meters.
Therefore, the water will strike the ground approximately 3.45 meters away from the base of the tank.