A body of density rho is dropped from rest at | vedclass

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(A) Let $V$ be the volume of the body. When the body hits the surface of the lake,its velocity $v_0$ is given by $v_0^2 = 2gh$.Inside the water,the fo

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$\frac{h \rho}{(\sigma - \rho)}$

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(A) Let $V$ be the volume of the body. When the body hits the surface of the lake,its velocity $v_0$ is given by $v_0^2 = 2gh$.Inside the water,the forces acting on the body are the weight $W = \rho V g$ (downward) and the buoyant force $F_B = \sigma V g$ (upward).The net upward force is $F_{net} = F_B - W = Vg(\sigma - \rho)$.The acceleration $a$ of the body inside the water is $a = \frac{F_{net}}{m} = \frac{Vg(\sigma - \rho)}{\rho V} = \frac{g(\sigma - \rho)}{\rho}$.Since the force is upward,the acceleration is directed upward (opposing the motion),so $a = -\frac{g(\sigma - \rho)}{\rho}$.Let $H$ be the maximum depth reached. At this depth,the final velocity $v = 0$.Using the equation of motion $v^2 - u^2 = 2aH$,we have:$0 - (\sqrt{2gh})^2 = 2 \left( -\frac{g(\sigma - \rho)}{\rho} \right) H$$-2gh = -\frac{2g(\sigma - \rho)}{\rho} H$$h = \frac{(\sigma - \rho)}{\rho} H$$H = \frac{h \rho}{(\sigma - \rho)}$.

$A$ body of density $\rho$ is dropped from rest at a height $h$ into a lake of density $\sigma$ $(\sigma > \rho)$. The maximum depth to which the body sinks before returning to float on the surface is (neglect air dissipative forces).

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