This is a question I came up with while watching my friend squirt ketchup onto his table. He was squirting the ketchup out of a bottle while moving the bottle upwards.

A ketchup bottle starts upside down with the tip at the table. Ketchup is squirted out at a volume flow rate of $Q(t)$, in $\frac{m^3}{s}$, while the bottle itself is moving upwards at a rate of $v(t$), in $\frac{m}{s}$.

When the ketchup comes out of the bottle it is always initially not moving, but immediately starts falling to the table due to gravity ($g = 10 \frac{m}{s^2}$ downwards).

Find $V(t)$, the volume of ketchup on the table as a function of time.

Edit 1: We can ignore the fact that in real life the accumulating ketchup on the table will, in some sense, increase the height of the table.

enter image description here

  • 3
    $\begingroup$ Thumbs up because of the bizarreness of your question :) $\endgroup$ – Gonzalo Benavides Jul 4 '18 at 2:27
  • 2
    $\begingroup$ +1 for the graphics! $\endgroup$ – JavaMan Jul 4 '18 at 2:30
  • 2
    $\begingroup$ I'd like to thank your friend for inspiring this absurd question. $\endgroup$ – Carser Jul 4 '18 at 2:30
  • 1
    $\begingroup$ Are you having to squeeze the bottle or is it flowing out naturally? $\endgroup$ – Mohammad Zuhair Khan Jul 4 '18 at 2:31
  • 2
    $\begingroup$ @MohammadZuhairKhan, right, having a known geometry would be fine. Although, I'm not sure that any given geometry is much more accurate than assuming no height... $\endgroup$ – Carser Jul 4 '18 at 2:39

If we let $x=0$ be the table top and measure upwards, the bottle position is $x(t)=vt$.
Ketchup that leaves the bottle at time $t$ takes $t'$ to fall where $vt=\frac 12g(t')^2$
Ketchup that leaves the bottle at time $t$ hits the table at $t+\sqrt{\frac {2vt}g}$
At time $u$ the ketchup that just arrived at the table left the bottle at time $t$ where $u=t+\sqrt{\frac {2vt}g}$
We would like to invert this equation $$u=t+\sqrt{\frac {2vt}g}\\(u-t)^2=\frac {2vt}g\\t^2-2ut-\frac {2v}gt+u^2=0\\ t=\frac 12\left(2u+\frac {2v}g-\sqrt{(2u+\frac {2v}g)^2-4u^2}\right)$$ and the amount of ketchup at time $u$ is $Q$ times this. If $Q$ is not a constant, you need to integrate $$\int_0^{t(u)}Q(t)dt$$to get the amount on the table at time $u$

  • 2
    $\begingroup$ This seems to be for a constant velocity $v(t)$, I think. What about when $v(t)$ is not constant? $\endgroup$ – Carser Jul 4 '18 at 3:32
  • 3
    $\begingroup$ @Carser: yes, I assumed constant $v$. If $v$ is not constant the same logic works. You need to look back in time to the time when the bottle was as a height that the ketchup that left is just hitting the table. If $v(t)$ is not simple you will have to find a numeric solution. $\endgroup$ – Ross Millikan Jul 4 '18 at 3:40

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.