Answer to Question #96978 in Mechanics | Relativity for Inyang

Question #96978
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;s_1=h=50 m;u=0;

Let v1v_1 be the velocity and t1t_1 be the time elapsed after 50m,so v1=2gh=2×9.8×50=31.30 m/sec50 m,so\ v_1=\sqrt{2gh}=\sqrt{2\times 9.8\times 50}=31.30\ m/sec

t1=2hg=2×509.8=3.194 sect_1=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 50}{9.8}}=3.194\ sec

During second phase:t2t_2  be the time taken,Initial velocity=v1=31.30 m/sec;a=2m/sec2=v_1=31.30 \ m/sec;a=-2m/sec^2

Final velocity(v2)=3m/sec(v_2)=3m/sec

    v2=v1+at2    3=31.302×t2    t2=14.15 sec\implies v_2=v_1+at_2\implies 3=31.30-2\times t_2\implies t_2=14.15 \ sec

Distance traveled with parachute or height at which parachute open=s2=v1t2+12at22;=s_2=v_1t_2+\frac{1}{2}at_2^2;

s2=31.30×14.15+12×(2)×14.152=s_2=31.30\times 14.15+\frac{1}{2}\times (-2)\times 14.15^2= 242.67m242.67 m

Hence (a)(a) Total time parachutist was in air=t1+t2=17.344 sec=t_1+t_2=17.344 \ sec

(b)(b) The height at which fall begin=s1+s2=292.67 m=s_1+s_2=292.67\ m


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