# Distance between two drops [closed]

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?

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

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• ​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. – Farcher Aug 21 at 10:28
• $d=2v_i\cos\theta_i$ is almost correct, but you want $d$ as a function of time, so you need to have $t$ somewhere in your expression. – gandalf61 Aug 21 at 10:46
• @gandalf61 yes I forgot the t, I meant $d=2v_icosθ_i\cdot t$ – user112112 Aug 21 at 14:19
• @Farcher thank you for bringing this to my attention – user112112 Aug 21 at 14:19

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.
• @NuclearWang No, it still will be a linear function of t! Actually it will be $d(t) = t * |\vec{V_1}-\vec{V_2}|$ – lesnik Aug 21 at 17:54