A cylindrical container with a height of 1.0 meters is filled with water. A small orifice is made at a height of 0.25 meters from the bottom. Calculate the horizontal range of the water stream flowing out of the orifice. Assume g = 9.8 m/s²

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We have a cylindrical container filled with water, standing 1.0 meters tall. An orifice is created at a height of 0.25 meters from the bottom. We need to find how far horizontally the water stream will travel when it flows out. First, we identify the key heights in our problem. The height of water above the orifice is h equals 1.0 minus 0.25, which gives us 0.75 meters. The vertical distance the water will fall from the orifice to the bottom is y equals 0.25 meters. These two measurements are crucial for calculating the horizontal range. Step 2 involves applying Torricelli's Law to find the horizontal velocity of the water stream. Torricelli's Law states that the velocity equals the square root of 2 times g times h. With g equals 9.8 meters per second squared and h equals 0.75 meters, we calculate v equals square root of 14.7, which gives us approximately 3.83 meters per second. Step 3 calculates the time it takes for the water to fall the vertical distance y. Using the vertical motion equation y equals one half g t squared, we solve for t to get t equals square root of 2y over g. With y equals 0.25 meters and g equals 9.8 meters per second squared, we find t equals approximately 0.226 seconds. Finally, we calculate the horizontal range using R equals v times t. With velocity 3.83 meters per second and time 0.226 seconds, we get R equals 0.866 meters. We can verify this using the alternative formula R equals 2 times the square root of h times y, which also gives us 0.866 meters. Therefore, the water stream travels horizontally 0.866 meters before hitting the ground.