A ball is released from the top of a tower of height h meters. It takes T seconds to reach the ground. What is the position of the ball in T /3 seconds
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\(\therefore \quad h - u + \frac{1}{2} g x^{2}\) \(= h - \frac{1}{2} g \tau^{2}\)
$h_2$ $t = T/3$ $h_2 - h'$ $h'$
After \(\frac{T}{3}\) seconds, the position of ball,
\(\hbar = 0 + \frac{1}{2} g \left| \frac{7}{3} \right|^{2} = \frac{1}{2} \times g^{9}, \tau^{2}\)
\(h' = - \frac{1}{2} \sqrt{\frac{g}{9}} \tau^{2}\) \(-\frac{11}{3}x\) from top
\ \ Position of ball from ground \(-h - \frac{h}{9} - \frac{8h}{9} m.\)
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