JEE Main 2005 Physics question
A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 . He reaches the ground with a speed of 3 m/s. At what height, did he bail out?
Answer: (3)
Solution
After 50 m of free fall, . With deceleration 2 m/s²: $ = 980 - 2×2× h ⇒ h =
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