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?
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
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.