A ball of relative density 0.8 falls into wat | vedclass

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(A) Let the density of the ball be $\rho_b = 0.8 \rho_w$ and the density of water be $\rho_w$. The height from which it falls is $h = 2 \ m$.When the

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) Let the density of the ball be $ho_b = 0.8 ho_w$ and the density of water be $ho_w$. The height from which it falls is $h = 2 \ m$.When the ball hits the water surface,its velocity $v$ is given by $v^2 = 2gh = 2 	imes g 	imes 2 = 4g$.Inside the water,the forces acting on the ball are gravity ($mg$ downwards) and buoyancy ($F_B = V ho_w g$ upwards).The net force $F_{net} = F_B - mg = V ho_w g - (V ho_b) g = V g (ho_w - 0.8 ho_w) = 0.2 V ho_w g$.The mass of the ball is $m = V ho_b = 0.8 V ho_w$.The retardation $a$ inside the water is $a = \frac{F_{net}}{m} = \frac{0.2 V ho_w g}{0.8 V ho_w} = \frac{0.2}{0.8} g = \frac{g}{4}$ (upwards).Using the equation of motion $v^2 = 2as$,where $v^2 = 4g$ and $a = -g/4$ (since it is retardation):$0 = v^2 - 2as_{depth} \Rightarrow 4g = 2 	imes (g/4) 	imes s_{depth}$.$4g = \frac{g}{2} 	imes s_{depth} \Rightarrow s_{depth} = 8 \ m$.

$A$ ball of relative density $0.8$ falls into water from a height of $2 \ m$. The depth to which the ball will sink is ........ $m$ (neglect viscous forces):

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