If the air resistance causes a vertical retardation of 10 percent of value of acceleration due to gravity, then the time of flight of a projectile will be decreased by nearly? Take $g= 10 \frac{m}{s^2}$

My attempt: Only the vertical component of the projectile will be affected by changes in $g$ so by doing the time of flight analysis for a particle thrown up with velocity equal to the vertical component of the velocity of projectile, the analysis will become easier.

I understand that air resistance is like friction; it acts in the direction opposite to the motion. So, when the particle is thrown up, the air resistance together with gravity will act in downward direction. According to the question, value of retardation due to air resistance is $1 \frac{m}{s^2}$. So, when the ball is thrown up, net acceleration equals $10+1$=$11 \frac{m}{s^2}$ in the downward direction.

When the particle is coming down after reaching its maximum height, air resistance should oppose its motion and act in vertically upward direction. Therefore, net acceleration must be $10-1$=$9\frac{m}{s^2}$ in the downward direction.

Time of flight=$\frac{u}{9}+\frac{u}{11}$ where u is the velocity of projectile in the vertical direction.

In the absence of air resistance, time of flight=$\frac{2u}{g}$=$\frac{u}{5}$

I think this analysis is wrong because it is not leading me to the answer. Where am I wrong?

  • $\begingroup$ Welcome to Physics SE! Such a well formatted, within-guidelines question from a new user? Must be a dream. Keep it up! $\endgroup$ Jun 23, 2017 at 12:51
  • $\begingroup$ @HritikNarayan I was on this site some time ago but then I lost access to that account. So, I know the rules and guidelines here. $\endgroup$ Jun 23, 2017 at 13:00
  • 3
    $\begingroup$ I don't think that the highlighted question makes much sense because I don't think that it is valid even as a first approximation to model the effects of drag by adjusting the effective gravitational acceleration acting on the object. Drag is a function of the velocity, and that basic point is completely ignored using the above approach. Where did you get this highlighted question? $\endgroup$
    – user93237
    Jun 23, 2017 at 16:54
  • $\begingroup$ @SamuelWeir This question was in some book I was solving. By "drag is a function of velocity", do you mean that air drag acting on the particle will keep on changing as the velocity of the particle changes? Why would that be? When we consider a block on a surface with friction, we say that after it overcomes the static friction, kinetic friction acts on it which is constant. The block might change its velocity due to the application of some external force upon it, but the kinetic frictional force remains constant. I understand air drag as a kind of friction only, where's the difference? $\endgroup$ Jun 24, 2017 at 8:37
  • 1
    $\begingroup$ @elle Air drag is usually modeled as $F_{drag}=-\beta v$ where $\beta$ is some drag coefficient, and $v$ is the velocity. $\endgroup$ Jun 24, 2017 at 9:53

1 Answer 1


Hint: I think the issue here is that the motion here is not symmetric. If it starts out with a speed $u$, it is not necessary that it will have the same speed when it reaches the bottom - because the acceleration is not the same in both the cases.

In the case without air resistance, it is valid to write $t=\frac{u}{g}+\frac{u}{g}$, because the particle goes from $u$ to $0$ and then $0$ to $u$, which doesn't happen in this case.

Try studying the problem by taking into account the distance the projectile travels - because the distance it travels up is always the distance it'll travel down.

  • $\begingroup$ @Hrithik Narayan On solving the question by taking into account the distance, I am getting $t=\frac{u}{11}+\frac{u}{3(11)^{1/2}}$, this still does not lead me to the answer. Will you be kind enough to check if I have arrived at the right result? I have rechecked my calculations a dozen of times. $\endgroup$ Jun 24, 2017 at 8:55
  • $\begingroup$ I am getting the final answer as 5 % whereas the answer given is 9%. $\endgroup$ Jun 24, 2017 at 9:10
  • $\begingroup$ Yeah I seem to be getting an answer of $4.2%$, I have a feeling the given answer might be wrong? $\endgroup$ Jun 24, 2017 at 9:58
  • $\begingroup$ If, however, we make the stupid mistake of assuming the acceleration to be $11$ in both cases, the answer seems to be close to $10$, so I think the question is wrong. $\endgroup$ Jun 24, 2017 at 10:10
  • $\begingroup$ Are you getting the time of flight when the ball goes up as $\frac{u}{11}$ and the time when the ball comes down as $\frac{u}{3(11)^{1/2}}$ ? $\endgroup$ Jun 24, 2017 at 10:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.