A stream-lined falls through air from a height $h$  on the surface of a liquid. Let $d$  and $D$  denote the densities of the materials of the body and the liquid respectively. if $D>d$ , then the time after which the body will be instantaneously at rest, is:

Velocity $u$  of the body when it enters the liquid is given by

$mgh=\frac{1}{2}m{u}^2$  or $u=\sqrt{2gh}$

Let volume of the body$=V$

Mass of the body$=Vd$

Weight of the body$=Vdg$

Mass of liquid displaced$=VD$

Weight of liquid displaced$=VDg$

Net upward force$=VDg-Vdg$

$=Vg\left(D-d\right)$

$\text{Retardation}\text{=}\frac{\text{net weight}}{\text{mass}}$

$=\frac{V\left(D-d\right)g}{Vd}=\left(\frac{D-d}{d}\right)g$

Acceleration $a=-\left(\frac{D-d}{d}\right)g$

Final velocity, $v$  in the liquid when the body is instantaneously at rest is zero. let the time taken be $t$ .

Now $v=u+at$

$0=\sqrt{2gh}-\left(\frac{D-d}{d}\right)gt$ , $\left(\frac{D-d}{d}\right)gt=\sqrt{2gh}$

$\therefore$  $t=\left[\frac{d}{D-d}\right]\sqrt{\frac{2h}{g}}$