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(D) The initial kinetic energy of the body is given by $K_i = \frac{1}{2} m u^2 = \frac{1}{2} 	imes 1 	imes (20)^2 = 200\, J$.If there were no air f

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$20$

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(D) The initial kinetic energy of the body is given by $K_i = \frac{1}{2} m u^2 = \frac{1}{2} 	imes 1 	imes (20)^2 = 200\, J$.If there were no air friction,the maximum height $H$ reached by the body would be determined by the conservation of energy: $m g H = K_i$.$1 	imes 10 	imes H = 200 \implies H = 20\, m$.However,the body only reaches a height of $h' = 18\, m$ due to energy loss from air friction.The potential energy at this height is $U_f = m g h' = 1 	imes 10 	imes 18 = 180\, J$.The energy lost due to air friction is the difference between the initial kinetic energy and the final potential energy: $\Delta E = K_i - U_f = 200\, J - 180\, J = 20\, J$.

$A$ body of mass $1\, kg$ is thrown upwards with a velocity $20\, m/s$. It momentarily comes to rest after attaining a height of $18\, m$. How much energy is lost due to air friction? $(g = 10\, m/s^2)$

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