AIIMS2019Physics-Projectile Motion

AIIMS 2019 Physics Horizontal Range MCQ Question

Type: MCQ-numerical-Medium-Class 11

The initial velocity of the water stream is given as, v=2g(Hh)v = \sqrt{2g(H⁻h)} The time taken by stream to hit the surface is, h = rac{1}{2}gt^2 t = \sqrt{ rac{2h}{g}} The horizontal range can be calculated as, Horizontal range = vt = \sqrt{2g(H⁻h)} imes \sqrt{ rac{2h}{g}} = 2\sqrt{(H⁻h)h} = 2\sqrt{(1-0.25)(0.25)} = 0.866 ext{ m}

Correct Answer

Option D

Detailed Explanation

The calculation of the horizontal range of the water stream is based on the derived formula Horizontal range=2(Hh)h\text{Horizontal range} = 2\sqrt{(H - h)h}, where HH is the initial height and hh is the height at which the stream is measured. Substituting H=1H = 1 m and h=0.25h = 0.25 m yields a range of 0.8660.866 m, confirming that the derived value aligns with the physics of projectile motion. Since the other options are not provided with any numerical values or context, they cannot be considered correct or relevant to the question. This exercise illustrates the application of kinematic equations in determining the trajectory of a fluid under the influence of gravity, emphasizing the importance of understanding the relationship between height, time, and horizontal distance in projectile motion.

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