= Solution
The air fall lasts
$$
t_a=\sqrt{\frac{2H_a}{g}}\simeq1.0096\,\mathrm s.
$$
Retain the underwater deceleration rather than replacing the whole water trajectory with motion at <terminal velocity>. Define
$$
\chi=\operatorname{arctanh}\frac{v_t}{u}>0,\qquad \nu=bv_t.
$$
Separation of the descent equation gives the <hyperbolic cotangent> solution and its integrated depth:
$$
v(t)=v_t\coth(\chi+\nu t),\qquad
z(t)=\frac1b\ln\frac{\sinh(\chi+\nu t)}{\sinh\chi},
$$
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> yields
$$
\boxed{t_w=\frac{\operatorname{arsinh}(e^{bH_w}\sinh\chi)-\chi}{bv_t}\simeq4.4264\,\mathrm s.}
$$
The total elapsed <time> is therefore
$$
\boxed{t_{\mathrm{total}}=t_a+t_w\simeq5.44\,\mathrm s.}
$$
For an estimate without evaluating the exact expression, $bH_w\gg1$ permits $\operatorname{arsinh}x\simeq\ln(2x)$ here, giving
$$
t_w\simeq\frac{H_w}{v_t}+\frac{\ln(2\sinh\chi)-\chi}{bv_t}.
$$
This gives the same $4.43\,\mathrm s$ to the quoted precision. The simpler $H_w/v_t\simeq4.72\,\mathrm s$ overestimates the underwater <time> because the pebble initially moves faster than $v_t$.
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