Question

A parachutist after balling out falls 50m without friction . when parachute opens , it decelerates at 2m/s² .He reaches the ground with a speed of 3m/s . At what height, did he bail out ? [AIEEE 2005] ​

Answer 1

Explanation :- Parachute bails out at height H from ground velocity at A

• V = √2gh
• √2*9.8×50 = √980 m/s
• Velocity at ground , V₁= 3m/s
• acceleration= -2m/s²
• h = 50m
• ∴ H - h = V² - V₁² /2×2
• H - h = (√980)² - 9/4
• H - h = 971/4
• H = 991 /4 + 50
• H = 242.75 + 50 ≈ 293 m Answer .

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