Show that total energy of a freely falling body remains constant during its motion.

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Solution

Let the energy of particle at any height = $E$
Initial energy = $\frac{1}{2} m v^{2} + m g H$
$= 0 + m g H = E_{o}$
Velocity at any height = $v$
$∴ v^{2} = u^{2} + 2 a s$
$v^{2} = 0 + 2 (- g) [- (H - h)]$
$= 2 g (H - h)$
$∴ E = \frac{1}{2} m v^{2} + m g h$
$= \frac{m}{2} . 2 g (H - h) + m g h = m g H = E_{o}$
Hence constant

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