Terminal Velocity of 100kg Parachutist Falling from 3000m

AI Thread Summary
The discussion focuses on calculating the terminal velocity of a 100kg parachutist falling from 3000m, considering air resistance. The parachutist experiences free fall for 30 seconds before the parachute opens, with air resistance modeled as proportional to velocity. The terminal velocity is defined as the point where the forces of gravity and air resistance balance, resulting in zero acceleration. The calculation provided yields a terminal velocity of 9.81 m/s after the parachute opens, which is confirmed as correct by other participants. Overall, the conversation emphasizes understanding the relationship between gravitational force and air resistance in determining terminal velocity.
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Homework Statement



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?

Homework Equations



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)



The Attempt at a Solution



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


i think this is wrong... but i don't know where to start... i think my concept on terminal velocity is wrong... can anybody help me... please3
 
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Your understanding is correct. The terminal velocity is the velocity at which the sum of the drag and gravity force equals zero.
 
Filip Larsen said:
Your understanding is correct. The terminal velocity is the velocity at which the sum of the drag and gravity force equals zero.

ok! then, is my understand is reflected by the correct calculations? help me to clarify this thing...
 
Yes, your calculation is also correct.
 
Filip Larsen said:
Yes, your calculation is also correct.

thanx 4 d clarification... rely appreciated it...:smile:
 
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