A bullet of mass 200,g having initial kinetic | vedclass

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(A) Given: Mass $m = 200\,g = 0.2\,kg$,Initial kinetic energy $K_i = 90\,J$,Final kinetic energy $K_f = 40\,J$ at time $t = 1\,s$.Using the relation $

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$45$

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(A) Given: Mass $m = 200\,g = 0.2\,kg$,Initial kinetic energy $K_i = 90\,J$,Final kinetic energy $K_f = 40\,J$ at time $t = 1\,s$.Using the relation $K = \frac{1}{2}mv^2$,we find the velocities:Initial velocity $u = \sqrt{\frac{2K_i}{m}} = \sqrt{\frac{2 	imes 90}{0.2}} = \sqrt{900} = 30\,m/s$.Final velocity $v = \sqrt{\frac{2K_f}{m}} = \sqrt{\frac{2 	imes 40}{0.2}} = \sqrt{400} = 20\,m/s$.Assuming constant retardation,the acceleration $a$ is:$a = \frac{v - u}{t} = \frac{20 - 30}{1} = -10\,m/s^2$.To find the distance $s$ required to come to rest $(v_{final} = 0)$:Using $v_{final}^2 = u^2 + 2as$,we get $0 = (30)^2 + 2(-10)s$.$20s = 900 \implies s = 45\,m$.

$A$ bullet of mass $200\,g$ having initial kinetic energy $90\,J$ is shot inside a long swimming pool as shown in the figure. If its kinetic energy reduces to $40\,J$ within $1\,s$,the minimum length of the pool the bullet has to travel so that it completely comes to rest is $.....m$.

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