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(B) Let the total height be $H = 30 \ m$ and the velocity of the body be $v$ at a point where potential energy $(PE)$ is twice the kinetic energy $(KE

Vedclass · Accepted Answer

$10 \sqrt{2} \ m \ s^{-1}$

Vedclass · Answer

(B) Let the total height be $H = 30 \ m$ and the velocity of the body be $v$ at a point where potential energy $(PE)$ is twice the kinetic energy $(KE)$.According to the law of conservation of energy,the total energy at any point is equal to the initial potential energy:$m g H = PE + KE$Given that $PE = 2 	imes KE$,we substitute this into the equation:$m g H = 2 	imes KE + KE = 3 	imes KE$Since $KE = \frac{1}{2} m v^2$,we have:$m g H = 3 	imes (\frac{1}{2} m v^2)$$g H = \frac{3}{2} v^2$$v^2 = \frac{2}{3} g H$Substituting the values $g = 10 \ m \ s^{-2}$ and $H = 30 \ m$:$v^2 = \frac{2}{3} 	imes 10 	imes 30$$v^2 = 2 	imes 10 	imes 10 = 200$$v = \sqrt{200} = 10 \sqrt{2} \ m \ s^{-1}$

$A$ body is released from a height of $30 \ m$ vertically downwards. The speed of the body at which potential energy is twice that of kinetic energy is (Acceleration due to gravity $= 10 \ m \ s^{-2}$)

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