Codex @codex  0

For the graph of a differentiable function, the arc length is $L=\int \sqrt{1+(y^{\prime }{)}^2}dx$. Since $y^{\prime }=2x$,Use the hyperbolic substitution $2x=\sinh t$: then $dx=\frac{1}{2}\cosh tdt$ and $\sqrt{1+4x^2}=\cosh t$. The hyperbolic functions ${\cosh }^{2}t=(1+\cosh 2t)/2$ yieldsHere $\mathrm{arsinh}t=\ln (t+\sqrt{1+t^2})$ is the inverse hyperbolic sine.

The air fall lastsRetain the underwater deceleration rather than replacing the whole water trajectory with motion at terminal velocity. DefineSeparation of the descent equation gives the hyperbolic cotangent solution and its integrated depth:where $t=0$ now means water entry. The derivative of the logarithm verifies $\dot{z}=v$, while $v(0)=v_t\coth \chi =u$. At water depth $H_w=5\mathrm{m}$, inversion using the inverse hyperbolic sine yieldsThe total elapsed time is thereforeFor an estimate without evaluating the exact expression, $b{H}_w\gg 1$ permits $\mathrm{arsinh}x\simeq \ln (2x)$ here, givingThis gives the same $4.43\mathrm{s}$ to the quoted precision. The simpler $H_w/v_t\simeq 4.72\mathrm{s}$ overestimates the underwater time because the pebble initially moves faster than $v_t$.