Finding Terminal Velocity given Mass and magnitude

AI Thread Summary
To find the terminal velocity of a 70 kg skydiver experiencing a drag force described by Fd(v)=0.19v^2, the gravitational force is calculated as 686.7 N using F=ma. At terminal velocity, the drag force equals the gravitational force, leading to the equation 686.7 N = 0.19v^2. Solving this gives a terminal velocity of approximately 60.1 m/s. The discussion emphasizes the importance of setting the magnitudes of the forces equal due to Newton's second law, particularly noting that acceleration is zero at terminal velocity. Understanding these principles is crucial for solving similar physics problems effectively.
Matt_h
Messages
7
Reaction score
0
I'm in physics 1 in college and reviewing for an up comming exam, ran into this problem and can't seem to figure out how to solve it.

A 70 kg skydiver experiences a drag force as a function of velocity (in m/s), opposite the direction of motion, whose magnitude is given

Fd(v)=0.19v^2

What is the skydiver's terminal velocity?

Any help would be appreciated!
 
Physics news on Phys.org
Hi, Matt h. Welcome to PF!

The rules here require you to show some attempt at a solution before receiving help. So, how do you think you could get started on this problem? What physics principles or laws are relevant?
 
Oh, ok

Using F=ma i found out that the force due to gravity is
70kg*9.81m/s^2=686.7N

Would i then just plug this force into the magnitude equation above to find the velocity?

After plugging it in i got 60.1 m/s

Does that sound about right?
 
Yes, that looks very good. Do you see why you should set the magnitudes of the two forces equal to each other?
 
  • Like
Likes 1 person
TSny said:
Yes, that looks very good. Do you see why you should set the magnitudes of the two forces equal to each other?

I'm not sure that i do, do you mind explaining it?
 
Newton's 2nd law: Fnet = ma

For an object moving at terminal velocity, what is the value of the acceleration a?
 
  • Like
Likes 1 person
Acceleration would be 0, right?

or no acceleration?
 
Yes. So,...:smile:
 
Back
Top