2
$\begingroup$

2 drops are thrown away simultaneously in $t=0$, with initial velocity of $v_i$ and in angle of $\theta_i$ in opposite directions as seen in the picture. What is the distance between them as a function of time?

This is a question from homework. I think the answer is $d=2v_i \cos\theta_i\cdot t$ but is seems too simple. Am I missing something?enter image description here

$\endgroup$

closed as off-topic by John Rennie, Jon Custer, ZeroTheHero, stafusa, Kyle Kanos Aug 23 at 10:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, Jon Custer, ZeroTheHero, stafusa, Kyle Kanos
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ ​Hi and welcome to the Physics SE! Please note that we don't answer homework or worked example type questions. Please see this page in the site help for more on what topics you can ask about here. $\endgroup$ – Farcher Aug 21 at 10:28
  • 1
    $\begingroup$ $d=2v_i\cos\theta_i$ is almost correct, but you want $d$ as a function of time, so you need to have $t$ somewhere in your expression. $\endgroup$ – gandalf61 Aug 21 at 10:46
  • $\begingroup$ @gandalf61 yes I forgot the t, I meant $d=2v_icosθ_i\cdot t$ $\endgroup$ – user112112 Aug 21 at 14:19
  • $\begingroup$ @Farcher thank you for bringing this to my attention $\endgroup$ – user112112 Aug 21 at 14:19
-1
$\begingroup$

Yes, the answer IS simple.

It would remain simple even if $\theta_i$ and magnitudes of velocities are different for these two drops!

To understand why is it so let's switch to the frame of reference which falls freely with $g$. Distance doesn't depend from frame of reference, right? In this frame of reference each drop moves with constant velocity (that is without acceleration). So, the distance between the drops would increase with constant rate and will be $d(t) = V * t$, where $V$ depends only from initial velocities of drops, but not from time.

$\endgroup$
  • $\begingroup$ It actually gets quite a bit more complicated if the angles or speed of the drops are not identical. The answer is so simple because there is no relative vertical motion between the drops, and the horizontal velocity of the drops is constant. If the angles or speeds are different from one another, the drops now have a relative vertical velocity that varies with time, so the distance between them is no longer a simple linear function of t. $\endgroup$ – Nuclear Wang Aug 21 at 15:55
  • $\begingroup$ @NuclearWang No, it still will be a linear function of t! Actually it will be $d(t) = t * |\vec{V_1}-\vec{V_2}|$ $\endgroup$ – lesnik Aug 21 at 17:54

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