A PARACHUTIST BAILS OUT AND FREELY FALLS 50M THEN THE PARACHUTE OPENS , AND THEREAFTER SHE DECELERATES AT 2M/S². SHE REACHES THE GROUND WITH A SPEED OF 3M/S.
1.) HOW LONG WAS THE PARACHUTE IN THE AIR?
2.) AT WHAT HEIGHT DID THE FALL BEGIN?
1
Expert's answer
2019-10-22T10:09:22-0400
There are two phases,first when she is without parachute for first 50 m and then with parachute.
First phase:s1=h=50m;u=0;
Let v1 be the velocity and t1 be the time elapsed after 50m,sov1=2gh=2×9.8×50=31.30m/sec
t1=g2h=9.82×50=3.194sec
During second phase:t2 be the time taken,Initial velocity=v1=31.30m/sec;a=−2m/sec2
Final velocity(v2)=3m/sec
⟹v2=v1+at2⟹3=31.30−2×t2⟹t2=14.15sec
Distance traveled with parachute or height at which parachute open=s2=v1t2+21at22;
s2=31.30×14.15+21×(−2)×14.152=242.67m
Hence (a) Total time parachutist was in air=t1+t2=17.344sec
(b) The height at which fall begin=s1+s2=292.67m
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