How much ketchup is on the table? (Ketchup flow rate problem) 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](

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 $u$