Terminal Velocity Let Suppose a bullet of 2 gram is falling with the speed of 200km/h.
Now how can we know the time after which it gains it terminal velocity.
Also when it gain its terminal velocity the force of friction and gravity will becomes equal and it stop accelarating do at this point the force of friction acting on it becomes constant or it go on increasing and also decrease the velocity of object.   
 A: It's not actually friction, it's the drag force, also called air resistance. Technically, they're different (even though people do sometimes say "air friction"). The drag force depends on the object's speed, shape, and size, as well as the density of the air, but in the ideal case you can assume that the object's shape and size and the density are constant.
In that case, when an object is falling at terminal velocity, the forces on it balance out, which means it has zero acceleration - in other words, its velocity doesn't change. And if the velocity doesn't change, then the drag force, which depends on velocity, won't change either. So the object is "stuck" in a steady state in which the drag force on it will remain constant as it falls.
In practice, though, an object never actually reaches terminal velocity unless it starts at terminal velocity. As it gets closer and closer to terminal velocity, the drag force on it gets closer and closer to exactly balancing out the gravitational force, which means that the net force on it, and thus its acceleration, becomes less and less. Basically, as the object gets closer to terminal velocity, the rate at which it continues to approach terminal velocity gets slower. It'll get continually closer to terminal velocity as time goes on, but it never quite reaches it.
Last year I wrote a blog post about this issue which discusses it using the actual math. It might be of interest to you.
A: If you assume that the atmosphere is constant (i.e. ignore the fact that density and pressure increase as you go downwards), then you could linearize the differential equation for the velocity versus time. [I.e. you expand the drag as a function of velocity about the terminal velocity, and keep the first order correction to drag]. If you were to do so, you will find that it exponentially approaches terminal velocity with some characteristic time.
In the more realistic case where density (and hence drag) is a function of altitude, you'll have to solve the 2nd order diffeq numerically.
A: Here is a link.
d/dt(v)=g-pv
plug in the values for g and p.
for the earths surface g=-9.8 m/s^2, and p depends on altitude/air density. 
d/dt(v)=-9.8-p*v
d/dt(v)=p*(v+(-9.8/p))
1/(v+(-9.8/p),dv=p,dt
integrate both sides to get
ln(v+(-9.8/p)=p*t+C
raising both sides as exponents of e gives
v+(-9.8/p)=e^(p*t+C)=C*e^(p*t+C)
v(t)=C*e^(p*t)-(-9.8/p)
if p is known, and the velocity at time=zero is known, then the following will yield the value of C;
v(0)=C*1-(-9.8/p)
You said (I think) that v(0)=-200km/hr=-55.56m/s
so: -55.56m/s=C*1-(9.8/p)        solve for C and plugin C into: v(t)=C*e^(p*t)-(-9.8/p)
C=-55.56+(9.8/p)
v(t)=(-55.56+(9.8/p))*e^(p*t)-(-9.8/p)     this equation is (after p is plugged in) complete
tv=g/p
set the v(t) equal to tv:
g/p=(-55.56+(9.8/p))*e^(p*t)-(-9.8/p)     solve this equation for t at which tv is attained
tv=terminal velocity
v=velocity
v(t)=velocity as a function of time
g=acceleration due to gravity
p=drag coefficient
t=time
C=constant
Termial velocity is a constant unless the altitude changes enough to change the drag coefficient p. If the altitude decreases, the drag coefficient (air resistance) will likely increase and thereby slow decread the tv since tv=t/p: as p increases tv decreases. 
