Explanation

(c) Second law of motion gives
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ h = ut + \frac{1}{2}gT^{2}$
or$\ \ \ \ \ \ \ \ \ \ \ h = 0 + \frac{1}{2}gT^{2}\ \ (\because u = 0)$
$\therefore\ \ \ \ \ \ \ \ \ \ \ T = \sqrt{\left( \frac{2h}{g} \right)}$
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At $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ t = \frac{T}{3}s,$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ s = 0 + \frac{1}{2}g\ \left( \frac{T}{3} \right)^{2}$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ s = \frac{1}{2}g\ .\frac{T^{2}}{9}$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ s = \frac{g}{18}\ \times \frac{2h}{g}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \because T = \sqrt{\frac{2h}{g}} \right)$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ s = \frac{h}{9}m\ $
Hence, the position of ball from the ground
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = h - \frac{h}{9} = \frac{8h}{9}m.$