Question

A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m/s
he reaches the ground with a speed of 3 m/s. At what height, did he bail out? (IIT - 2005)
a) 91 m
b) 182m
c) 293 m
d) 111
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Answer 1

Answer:

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Answer 2

Final velocity, v = +3 m/s

Acceleration, a = -2m/s² + 10m/s² = 8m/s²

When he bails out (fancy way of saying drops from the plane), he falls with 0 initial velocity. So, by the end of 50 m, he has the velocity √2gh = √2(10)(50) = 10√10 m/s

Now, his initial velocity for the opened parachute is 10√10m/s

2 (8) (s) = 3² - 1000

s = 60.68

So, the answer must be s+50 i.e, 110.68 which is nearest to 111m. The discrepancy occured due to value of g. Another discrepancy would be whether to consider the actual acceleration due to gravity when the parachute is open. Well, to adress that, the acceleration due to gravity hasn't disappeared yet, so it has to be considered.