X = (3000 + a*100) , M = (75 + c) Kg Cd = 0.5 (skydiver) (Area 0.4m2) Cd = 1.2 (

Question

X = (3000 + a*100) ,
M = (75 + c) Kg
Cd = 0.5 (skydiver) (Area 0.4m2)
Cd = 1.2 (parachute) (Area 40m2)

A sky diver, weighing M Kg, jumps from a moving aircraft at X meters above the ground. He/She initially free-falls for a period of seconds then deploys a parachute to to allow him/her to land at a safe speed of under 6 m/s.

The sky diver uses a spread-eagle position during freefall that maximises their drag coefficient , Cd. (Cd based on frontal area of 0.5m2) Full parachute deployment takes 5 seconds and during this period the drag coefficient of the parachute can be assumed to vary linearly between 0 and the final value needed for a safe descent speed. Using whatever numerical techniques appropriate determine a velocity versus time graph of his/her descent.

Estimate the amount of free fall time that would be possible that would still allow full deployment of the parachute and a safe landing. Ignore any initial horizontal velocity of the aircraft and wind or atmospheric turbulence conditions.

Explanation / Answer

1.

For the free fall,

u=0

v=u+gt

V= gt for a<t<=0

Since, freefall takes place for a seconds

V(final)=ga

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For the time after parachute opens,

v=ga

ro denotes density of air,

F= Force due to earths attraction- [Drag due to parachute and skydiver +Drag due to parachute]

F=mg-[Cd(0.5*ro*V^2)*S+Cd(0.5*ro*V^2)*S]

F=(75 + c)g-(0.5*ro*(ga)^2)[0.5*0.4+Cd*40]

Since, of parachute varies linearly with time,

Cd=1.2/5 *t=0.24t

F=(75 + c)g-(0.5*ro*(ga)^2)[0.2+0.24t*40]

F=(75 + c)g-(0.5*ro*(ga)^2)[0.2+0.24t*40]=(75 + c)g-(0.5*ro*(ga)^2)[0.2+9.6t]

a=F/m=g-{(0.5*ro*(ga)^2)*[0.2+9.6t]}/(75+c)

v=u+at

v=ga +gt-{(0.5*ro*(ga)^2)*[0.2+9.6t]*t}/(75+c) a<=t<a+5

v(final)=ga +5g-{(0.5*ro*(ga)^2)*[48.2]*5}/(75+c)

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For the time after the parachute has opened fully,

Since, velocity is constant

F=0

F=(75 + c)g-(0.5*ro*(ga)^2)[0.2+48]

(75 + c)g=(0.5*ro*(ga)^2)[0.2+48]

v= 6m/s t>a+5

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