Answer to Question #96978 in Mechanics | Relativity for Inyang

Question

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?

Answer 1

There are two phases,first when she is without parachute for first 50 m and then with parachute.

First phase:$s_1=h=50 m;u=0;$

Let $v_1$ be the velocity and $t_1$ be the time elapsed after $50 m,so\ v_1=\sqrt{2gh}=\sqrt{2\times 9.8\times 50}=31.30\ m/sec$

$t_1=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 50}{9.8}}=3.194\ sec$

During second phase:$t_2$  be the time taken,Initial velocity$=v_1=31.30 \ m/sec;a=-2m/sec^2$

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

$\implies v_2=v_1+at_2\implies 3=31.30-2\times t_2\implies t_2=14.15 \ sec$

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

$s_2=31.30\times 14.15+\frac{1}{2}\times (-2)\times 14.15^2=$ $242.67 m$

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

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