# terminal velocity

• Apr 20th 2010, 05:38 AM
bobey
terminal velocity
A parachutist whose mass is 100kg drops from rest to an altitude of 3000m and falls under of gravity. Assume that the force due to to air resistance is proportional to the velocity of the parachutist , with the proportionality constant k1 = 20kg/s when the chute is closed and k2=100 kg/s when the chute is opened. The chute does not open until after 30s of free fall (9.81 m/s^2)

what is the terminal velocity after the chute opens?

my attempt:

i found the definition of terminal velocity : The terminal velocity of a falling body occurs during free fall when a falling body experiences zero acceleration...

thus my understanding is this terminal velocity is when f=mg (falling depends on gravity only)

==> k2v=mg
v=mg/k2
= 100(9.81)/100 = 9.81 m/s (Shake)

i think this is wrong... but i dont know where to start... i think my concept on terminal velocity is wrong... can anybody help me... plz3x(Speechless)
• Apr 20th 2010, 05:52 PM
hollywood
I believe your answer is correct. The terminal velocity is when the force of gravity balances the force due to air resistance (and therefore acceleration is zero), and your calculations are correct. But you don't use a lot of the data in the problem (3000m, 20kg/s, 30s).

Could it be that the problem is (mis-)using the term "terminal velocity" to mean the velocity when the parachutist hits the ground? If so, then you need to solve a differential equation to get the position and velocity when the chute opens, then solve another (with k2 instead of k1) to find the velocity at the ground.

Post again if you're still having trouble.

- Hollywood