JEE Advanced 2021 Paper 1 · Q06 · Projectile Motion

Correct answer:7.50

Solution

Launch speed 5√2 m/s at 45° from vertical ⇒ horizontal component v_x = 5 m/s, vertical component v_y = 5 m/s. Time of flight to apex = 5/10 = 0.5 s; horizontal range to apex = 5 × 0.5 = 2.5 m. Half-range = 2.5 m, so x-coordinate of apex from O is 2.5 m. At apex, by conservation of horizontal momentum: m·5 = (m/2)·0 + (m/2)·v ⇒ v = 10 m/s for the second mass. Second mass takes 0.5 s to fall, traveling horizontally 10 × 0.5 = 5 m. Total distance from O = 2.5 + 5 = 7.5 m.

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