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5.Work, Energy, Power and Collision
medium
A body of mass $2\, kg$ is thrown up vertically with K.E. of $490$ joules. If the acceleration due to gravity is $9.8$$m/{s^2}$, then the height at which the K.E. of the body becomes half its original value is given by ............ $\mathrm{m}$
A
$50 $
B
$12.5$
C
$25$
D
$10$
Solution
(b)Let h is that height at which the kinetic energy of the body becomes half its original value i.e. half of its kinetic energy will convert into potential energy
$mgh = $ $\frac{{490}}{2}$
==> $2 \times 9.8 \times h = \frac{{490}}{2}$
==> $h = 12.5m.$
Standard 11
Physics
