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A projectile is launched at 100 m/s at 37 degrees above the horizontal. At the highest point, it splits into two fragments whose masses are in the ratio 1: 3. The lighter fragment instantly comes to rest. Find the horizontal distance (in m) from the launch point to where the heavier fragment lands. (g = 10 m/s², sin 37 = 0.6, cos 37 = 0.8)

1. 750
2. 1000
3. 1120
4. 1250

Correct answer: 1120

Solution

At the highest point the velocity is purely horizontal: vx = 100 * 0.8 = 80 m/s. Let total mass = 4m. By momentum conservation (horizontal): 4m * 80 = m * 0 + 3m * v2 → v2 = 320/3 m/s. Height at highest point: H = (100 * 0.6)² / (2 * 10) = 180 m. Time to fall: t = sqrt(2H/g) = sqrt(36) = 6 s. Horizontal distance to highest point: x1 = vx * (v sin theta / g) = 80 * (60/10) = 480 m. Extra distance of heavy piece: x2 = v2 * t = (320/3) * 6 = 640 m. Total = 480 + 640 = 1120 m.

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