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.


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.

  • $\begingroup$ @David - I think that the difficulty here is to explain (or show) why there is no overshoot of the velocity when approaching the Vt. Since this is a nonlinear equation, one needs to actually prove that the convergence is exponential (or similar). I think your blog post does it well. $\endgroup$ – ysap May 15 '11 at 5:12
  • $\begingroup$ @ysap: fair enough; it seemed to me based on the style of the question that the OP might not want to get into the full mathematical detail, so I opted for a less technical explanation (with a link to more mathematical detail for anyone interested, of course). $\endgroup$ – David Z May 15 '11 at 5:27
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    $\begingroup$ @ysap -- As long as you know the equation is second-order -- that is, that the force doesn't depend on higher-order derivatives than the acceleration -- there can't be any overshoot. By definition, the terminal velocity is the velocity such that $a=0$. So if at any point the velocity reaches that terminal velocity, then the acceleration at that moment will be zero, and the unique solution to the differential equation from then on will be $v=$ constant. Heuristically, if there were overshoot, it would mean that there was a moment when $v$ is the terminal velocity but $a>0$, which is impossible. $\endgroup$ – Ted Bunn May 15 '11 at 14:28
  • $\begingroup$ will you please explain why an object never actually reaches terminal velocity unless it starts at terminal velocity. $\endgroup$ – Zain May 15 '11 at 14:31
  • $\begingroup$ Zain: Under the constant atmosphere simplification, you need some excess of graviational force versus drag in order for the velocity to increase. As the velocity approaches terminal velocity the net acceleration approaches zero, so it takes an infinire amount of time to reach that state. In the real world, an object falling from a great height will reflect the terminal velocity of the less dense air some height above its current height, and could in fact exceed the locally defined terminal velocity. $\endgroup$ – Omega Centauri May 15 '11 at 14:42

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.


Here is a link.


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



integrate both sides to get


raising both sides as exponents of e gives



if p is known, and the velocity at time=zero is known, then the following will yield the value of C;


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)


v(t)=(-55.56+(9.8/p))*e^(p*t)-(-9.8/p) this equation is (after p is plugged in) complete


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.

  • $\begingroup$ @Joe - I'm not sure your first eqn. is correct, the drag is a function of v^2, not v. So, it should be dv/dt = g-pv^2. $\endgroup$ – ysap May 15 '11 at 4:23
  • $\begingroup$ @Joe - Also, where does the number -55.56m/s come from? Then, the unit for g is m/s^2 and not m/s. $\endgroup$ – ysap May 15 '11 at 4:26
  • $\begingroup$ 55.56m/s=200km/hr to keep the units consistent with acceleration due to gravity (9.8m/s). it is negative because gravity is down. Drag is a function of v^2. However the change in drag with respect to time is the derivative of that. d(v^2)/dt=2*v therefore d/dt(v)=~stuff~v^(1) $\endgroup$ – Dale May 15 '11 at 4:31
  • $\begingroup$ @Joe - sorry, but I don't follow. 1) Going from the 1st eqn to the 2nd one in the answer, you substituted -55.56 for g. 2) dv/dt has units of m/s^2, yet you equate it with -55.56 m/s. 3) why do you take the derivative of v^2 in your above comment? $\endgroup$ – ysap May 15 '11 at 4:42
  • $\begingroup$ It would be helpful if you could edit your answer and start at the initial (nonlinear) diff. eqn., then explain a little more the steps and substitutions you perform. $\endgroup$ – ysap May 15 '11 at 4:45

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