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I am looking at the solution of a physics/statistics hybrid exercise, and I can't figure out how one expression step took place.

I have that

${dx \over dt} = gt$

$T = \sqrt{2h \over g}$

where $g$ is gravity acceleration, $h$ is the height of free fall, $dx$ is a distance interval, $dt$ is the corresponding time interval and $T$ is the full duration of the fall. I want the probability of finding the free falling object at a given position interval when piking a random time during its trajectory. So it starts with the probability of picking some time interval $dt$ of the full duration time $T$:

${dt \over T} = {dx \over gt}\sqrt{{g \over 2h}}$

So far, so good, then comes the step I fail to understand:

${dx \over gt}\sqrt{{g \over 2h}} = {1 \over 2 \sqrt{hx}}dx$

Could someone please explain this step?

1 Answers 1

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They integrated the first equation to get $x=\frac 12 gt^2$, which is valid as long as $x(0)=0$. Then they multiplied by $1=\frac {t\sqrt g}{\sqrt {2x}}$