A body of mass 2, kg is thrown up vertically | vedclass

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Question

(B) The initial kinetic energy $(K_i)$ is $490\, J$. At the height $h$ where the kinetic energy becomes half of its original value,the kinetic energy

Vedclass · Accepted Answer

$12.5$

Vedclass · Answer

(B) The initial kinetic energy $(K_i)$ is $490\, J$. At the height $h$ where the kinetic energy becomes half of its original value,the kinetic energy $(K_f)$ is $\frac{490}{2} = 245\, J$. According to the law of conservation of energy,the loss in kinetic energy is equal to the gain in potential energy $(U)$. Loss in $K.E. = K_i - K_f = 490 - 245 = 245\, J$. Gain in potential energy $U = mgh$. Equating the two: $mgh = 245$. Substituting the given values: $2 	imes 9.8 	imes h = 245$. $19.6 	imes h = 245$. $h = \frac{245}{19.6} = 12.5\, m$.

$A$ body of mass $2\, kg$ is thrown up vertically with a kinetic energy of $490\, J$. If the acceleration due to gravity is $9.8\, m/s^2$,then the height at which the kinetic energy of the body becomes half its original value is ............ $m$.

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