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A skydiver jumps out and drops freely through 50 m, with air resistance negligible. After the parachute opens, he slows down with a constant deceleration of 2 m/s². If his speed on reaching the ground is 3 m/s, from what height did he jump? [2005]

1. 182 m
2. 91 m
3. 111 m
4. 293 m

Correct answer: 293 m

Solution

After free fall of 50 m: v^2 = 2g*50 ~ 980-1000. Then decelerating at 2 m/s^2 to 3 m/s: s = (v^2 - 9)/4 ~ 243 m. Total height ~ 50 + 243 = 293 m.

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