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I should know how to do this but I don't. I'm not very familiar with vectors. Perhaps I will be after this.

So I have a stream of water falling out of a pipe. It obviously forms a parabola of the form $f(x) = -ax^2+h$, where $h$ is the height of the pipe from the ground, $a$ is some constant and $x$ is the horizontal distance from the pipe.

The question is: if I know $h$, and I know at what distance the water hits the ground, what is the initial speed of the water as it exits the pipe, assuming that it exits the pipe parallel to the ground?

I'm going to try to figure this out myself, but I'd like to return to see how you guys solve this. (btw, not a homework question. I'm 30 years old and just curious.)

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If the water exits horizontally (you seem to assume that) the height of the water is $y=h-\frac {gt^2}2$. The time to reach the ground comes from setting $y=0, t=\sqrt {\frac{2h}{g}}$The horizontal position is $x=vt$ where $v$ is the velocity on exit from the pipe. Let $d$ be the horizontal distance where the water hits the ground. We have $v=\frac dt=d\sqrt{\frac g{2h}}$. Your $a$ is my $g$, the acceleration of gravity, about $9.8 m/s^2$ or $32 ft/s^2$.

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  • $\begingroup$ That's how I solved it too. Thanks! $\endgroup$ May 7, 2012 at 14:44

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