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What happens when a parachutist after bailing out falls for 10 seconds without friction when the parachute opens he descends with an acceleration of 2m/s^2 against his direction and reach the ground with 4m/s then find from what height he has dropped off ? Pls clear my confusion that is when a parachutist falls under an acceleration do the net acceleration becomes (10-2)m/s^2 and if not why not.....??? Pls explain thoroughly

What happens when a parachutist after bailing out falls for 10 seconds without friction when the parachute opens he descends with an acceleration of 2m/s^2 against his direction and reach the ground with 4m/s  then find from what height he has dropped off ?                                                  
 Pls clear my confusion that is when a parachutist falls under an acceleration  do the net acceleration becomes (10-2)m/s^2 and if not why not.....???  Pls explain thoroughly 

Grade:11

5 Answers

Jagrati
10 Points
5 years ago
This is me who asked the question want to notify that anser is 2996m.....Therefore pls answer  my question its very urgent pls anybody.
 
Venkat
273 Points
5 years ago
For a free falling object the velocity is
v = gt = 10*10 = 100 m/sec
 
For this velocity the height covered will be calculated as,
v2 = 2gh → h =sqrt(1002/(2*10)) = 500 m
 
He reaches ground with speed, v=4m/s with deceleration a=-2m/s². 
Then, v²-u²=2(a.s) → s=(42-1002)/(2*(-2))=9984/4 =2,496 m.
The height at which he bailed out would be h = 2,496+500 = 2,996 m.
Venkat
273 Points
5 years ago
there is typo error in above probelm. In fourth line read remove (sqrt) from the line. the corredt form is ,
/(2*10)) = 500 m2 = 2gh → h =(1002v
Venkat
273 Points
5 years ago
Sorry something messed up
 
there is typo error in above probelm. In fourth line read remove (sqrt) from the line. the corredt form is ,
/(2*10)) = 500 m2 = 2gh → h =(1002v
Jagrati
10 Points
5 years ago
Hey pls explain thoroughly what i asked that why don't we calculate net acceleration.

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