The problem statement is as follows:

Two balls of masses $M$ and $2M$ are thrown horizontally with the same initial velocity $u$ from the top of a tall tower and experience a viscous drag of $-kv$ ($k>0$) where $v$ is the instantaneous velocity. Compare the ranges of the two projectiles.

Now, I separately considered the motion of a ball horizontally and vertically. The only component of acceleration along the horizontal direction was provided by drag. So the equation would be , a = -kp/m( a is the horizontal acceleration, and p is the instantaneous horizontal velocity. ) The acceleration is inversely proportional to the mass and hence the heavier ball would have lesser acceleration, and then the heavier ball would hit the ground further away. My instincts directed me to integrate twice as follows: p dp/dx = -kp/m ( x is the horizontally displacement) and then integrating w.r.t time(t), this is the equation I obtained: ln(x) = -kt/m. Now I am second guessing myself because I have a feeling that the time of flight wouldn't be equal for both the masses. I tried to calculate the time of flight by analysing the vertical motion and obtained a differential equation I am finding hard to solve.(I'm just a high school student who's not yet done with half of the course's calculus part :P) Was my first solution correct? Or will successive integration give me a different answer altogether?

  • 2
    $\begingroup$ Do note that $a_x(t) = -\frac{k}{m}v_x(t)$, and is therefore not a constant depending on the instantaneous horizontal velocity $p$. To solve this you do need to solve a DE, but luckily the most trivial form: $\ddot{x}(t) = -\frac{k}{m}\dot{x}(t)$, with $\dot{x}(0) = p$ and $x(0) = 0$. The resulting solution is likely some exponential function like $x(t) = -\frac{pm}{k}e^{-\frac{k}{m}t}+p$. Now its up to you to do the same for the vertical direction, which is important to determine when the ball has hit the ground! $\endgroup$
    – Petrus1904
    Nov 16 '20 at 15:29
  • $\begingroup$ Doesn't the drag exist in the vertical direction also? $\endgroup$ Nov 16 '20 at 16:02
  • $\begingroup$ When you integrated, you omitted the. constant of integration. The differential equation for x should read: $$\frac{dx}{dt}=u-\frac{k}{m}x$$ $\endgroup$ Nov 16 '20 at 16:24
  • $\begingroup$ @Petrus1904 So indeed, I will not be able to answer it accurately till I solve the DE obtained in the case of vertical motion too. Thank you:) $\endgroup$ Nov 17 '20 at 16:09
  • $\begingroup$ @Chet Miller Oh yes, noticed that. But will that change anything? $\endgroup$ Nov 17 '20 at 16:11

I haven't seen your final solution for the horizontal direction. I get $$x=\frac{mu}{k}\left(1-e^{-\frac{k}{m}t}\right)$$I would have started analyzing the horizontal direction differently, by first solving for the velocity as a function of time: $$\frac{dv_x}{dt}=--\frac{k}{m}v_x$$Subject to the initial condition $v_x=u$ at t = 0, this integrates to $$v_x=ue^{-\frac{k}{m}t}$$Then you have $$\frac{dx}{dt}=ue^{-\frac{k}{m}t}$$This integrates to the (correct) result I gave above for x.

  • $\begingroup$ Thanks for that correction, although it would be a real help if you could solve the equation for the motion in the vertical direction as well. $\endgroup$ Nov 18 '20 at 8:08
  • $\begingroup$ How would that help you solve this problem? (All the information you need to solve this problem is contained in the solution for the horizontal direction) $\endgroup$ Nov 18 '20 at 11:59
  • $\begingroup$ I'm not very sure about the time of their flight. They may not be equal. That needs to be calculated as well.(At least, that's what I think) $\endgroup$ Nov 18 '20 at 12:00
  • $\begingroup$ With drag present, if the mass is launched horizontally from a very high elevation, is there a limit to how far the mass can travel horizontally? (Check the solution for the horizontal distance as a function of time) $\endgroup$ Nov 18 '20 at 12:04
  • $\begingroup$ IOW, what is x at infinite time? $\endgroup$ Nov 18 '20 at 15:01

From Petrus' comment : $$a_x=\frac{dv_x}{dt}=\frac{dv_x}{dx}\frac{dx}{dt}=\frac{dv_x}{dx}v_x=-\frac{k}{m}v_x$$ Integrating gives you $x$ as a function of $v_x$. Setting $v_x \to 0$ gives you the range.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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