Free Fall with Air Resistance - Terminal Velocity (video 1 of 2) 

Yeah I agree. I think you don't need to factor or make the integrals definite because you can solve for C at the end is a lot easier.

@@andrewkurtz1252 I think you have to factor to get v with dv so that you can take the integral on both sides.

Just subtite a with dv/dt and separate the integrals and solve both the intregals, your intial conditions will be your limits.

Just integrate from the initial velocity instead of from 0, 0 in this case is the initial velocity.

If instead of g we have an applied force F then the terminal velocity will depend on applied force. The terminal velocity will always be obtained when we balance forces ( in this case gravity and drag). Be careful when you say “terminal velocity depends on acceleration). Once terminal velocity is reached the acceleration goes to 0.

@@PhysicsNinja Yeah, I meant force. I always associate force with acceleration, thus the confusion. Thanks, for the reply. I just wanted to make sure the drag force can arise due to any kind of movement of the object and it's not somehow bound to the G force.

@@PhysicsNinja Yeah, I meant force. I always associate force with acceleration, thus the confusion. Thanks, for the reply. I just wanted to make sure the drag force can arise due to any kind of movement of the object and it's not somehow bound to the G force.

@@PhysicsNinja does that mean we can control terminal velocity, in horizontal direction also, if we control the horizontal force?

@@bandekarameya Yes! For projectile motion the drag force is always opposite of the instantaneous velocity so there are 2 components.

That’s an advanced problem. Gravity, buoyancy, drag, hmmm I’ll have to think about that one.

@@PhysicsNinja it's nice to make videos about it

Sam D yes should be dt but these rest looks good.

I think he's defined the upper limit as t' to differentiate it from the time function itself, just for clarity's sake

Where did you find those?

If we have the expression for v(t). Say you want to know how long it takes to reach 90% of vterm. Set v(t)=0.9*v_term and solve for t. Time will be the only unknown assuming you know b.

@@AndyMBBallSkills Thank you, I always tried to use limit to figure out, but it always diverged and I wasn't sure how I could do further.

btw I found the derivative of the velocity formula with respect to time and evalueted it at t=0 and the slope is g at the beginning just as told but I am still not able to understand how e^(-bt/m) became 1-(bt/m) in short time limit.

That is the Taylor expansion for exp(-a) what a is small. The expansion is 1-a+a^2/2-....I only keep the first 2 terms of the expansion

@@PhysicsNinja thank you very much for your quick reply. You helped me quite a lot.

What I mean is, what if v0 is not 0. How do I find v(t) ?

@@leomadden7724 If you look when i integrate the equation for time i integrate from 0 to some time later t, my limits for velocity go from 0 to some final velocity. You need to redo the integration with new limits for velocity. For you case you are integrating from 8 (at time 0) to some final value v. After you get your expression you substitute t=10seconds

Marcos Pimentel check out my other video where I look at the c squared case. The drag forge dependence on velocity depends on the Reynolds numbers. For slow moving objects the drag forge depends on v and for faster moving it depends on v squared.

@@PhysicsNinja thanks

b is the density of the air * the area of the object facing the direction of movement / 2 softschools.com/formulas/physics/air_resistance_formula/85/

@@realpowergaming1449 hey, I don’t understand why in the video Fr=bv whereas on the link you sent it is F=kv^2 Why is it squared in one and not in the other?

@@oscarc.2127 man can't do the integral of 1/(v^2-mg/b) dv looooool

I wrote down the drag force as being proportional to the velocity of the object. In this case b is just the constant of proportionality. At the end of the day the force should have the proper units so b would need to be in units of [N/(m/s)]. It will depend on the geometry of the object and the density of the fluid. The exact expression can be found in books on fluid dynamics (i'm guessing). Depending on the speed of the object moving through the fluid sometimes the drag force is proportional to v^2 rather than just v. Again there is a constant term in from of the velocity term.

b is the fluid resistance constant....It depends on the density of air(ρ) , Area of falling body(A), and aerodynamic drag(C). b=(ρCA)/2.