Question

2. A 60 kilogram skydiver (more correctly termed a "skyflopper", as the picture below indicates) jumps from a plane at a height of 1000 meters with an initial velocity of v(0) = 0 meters per second, and falls subject to air resistance whose magnitude is given by 15e, for 25 seconds. At this time his parachute opens and the air resistance is now 180. Assume that the positive direction is downward and the gravitational acceleration constant is g -9.8 m/s, and describe fully how you derived the relevant differential equations. Use MAPLE to find his speed at impact, and the time at which that occurred, and discuss whether or not he made it safely to the ground. Note that it is known that survivability decreases markedly for fall heights above 4 meters or so, corresponding to a terminal velocity v2gh9 meters per second.

Answer 1

Initially when drive starts

Only force on diver is = Mg (since v(0) =0 m/s -Resistance =0)

As he accelerates down and picks up speed

Net Force = Mg - 15 I v I = Mg - 15 gt for 25 seconds

Velocity at time t = V(0) + $\int_{0}^{t}adt$ = 9.8 t- (15/60) ( 9.8 ) t^2/2 = (9.8 x 25) - (9.8/4) ( 25^2/2) =