Terminal Velocity of a Falling Parachutist

[Saptarshi Sarkar]

Homework Statement

A parachutist is falling with a speed of 55m/s when his parachute opens. The total weight of the man and the parachute is W and the air resistance is Wv²/25 N, where v is the instantaneous velocity. The terminal velocity is

A) 10m/s
B) 5m/s
C) 11m/s
D) zero

Relevant Equations

W d²y/dt² = Wg - Wv²/25

I tried to calculate it by the way I know, i.e., setting the right hand side of the equation of motion to zero and getting

v² = 25g = 2500 (taking g =10)
=> v = 50m/s

But this answer is incorrect. How do I use the information of the initial velocity and why would it effect the terminal velocity? Shouldn't the terminal velocity be the same no matter what the initial velocity was?

 

[TSny]

W is the weight, not the mass.

 

[Saptarshi Sarkar]

W is the weight, not the mass.

Thanks! I can't believe I did that mistake.

 

[jim421616]

Also, 25g = 250, not 2500.

 

[MattB]

This raises another question. Why would the air resistance depend upon W?

 

[jbriggs444]

This thread is about 3.5 years old. The original poster has not been seen for the last 1.5 years. We can probably ignore the problematic relationship of air resistance to weight.

We can also ignore the erroneout units of measurement specified in the problem statement. Velocity has units. It is not a pure number.

Clearly the problem is an exercise in formula manipulation, not physics.