A parachutist after bailing out falls 52m without friction when the parachute opens she decelerats at 2.1m/sec.sec and reach the ground with a speed of 2.9m/sec How long has been the parachutist in the air
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The answer is coming t = 9.88 = 10seconds (approx).
Approach the question using t=√(2h/g) to calculate time of first 52 m fall which comes out to be 4.56 seconds.
Next calculate the v or final velocity of 52m fall which will act as u or initial velocity for the fall in which the parachute is opened. This comes out to be 45 m/s . Do note that the resultant acceleration now will become (a-g) which is -7.9m/s•s .
Now putting values of v (2.9 as given) and u (45 as calculated) and a.res (a-g or -7.9 as calculated) in the relation v=u+(a.res*t) we get the value of t to be 5.32 seconds.
Adding both the value of time obtained in both calculations , we get 4.56 + 5.32 = 9.88 seconds which can be rounded off to 10 seconds.
Hit thanks if helped.
Approach the question using t=√(2h/g) to calculate time of first 52 m fall which comes out to be 4.56 seconds.
Next calculate the v or final velocity of 52m fall which will act as u or initial velocity for the fall in which the parachute is opened. This comes out to be 45 m/s . Do note that the resultant acceleration now will become (a-g) which is -7.9m/s•s .
Now putting values of v (2.9 as given) and u (45 as calculated) and a.res (a-g or -7.9 as calculated) in the relation v=u+(a.res*t) we get the value of t to be 5.32 seconds.
Adding both the value of time obtained in both calculations , we get 4.56 + 5.32 = 9.88 seconds which can be rounded off to 10 seconds.
Hit thanks if helped.
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