A ball is released from the top of a tower of height h metres. It takes T seconds to reach the ground. What is the position of the ball in T/3 seconds?

h/9 metres from the ground

7h/9 metres from the ground

8h/9 metres from the ground

17h/18 metres from the ground.

Solution

8h/9 metres from the ground

second law of motion

straight s space equals ut space plus space 1 half space plus gT squared or space straight h equals 0 plus 1 half gT squared space left parenthesis because space straight u equals 0 right parenthesis therefore space straight T equals square root of open parentheses fraction numerator 2 straight h over denominator straight g end fraction close parentheses end root At space straight t space equals space straight T over 3 straight s comma straight s space equals space 0 plus space 1 half straight g space open parentheses straight T over 3 close parentheses squared rightwards double arrow space straight s space equals space 1 half space straight g. straight T squared over 9 rightwards double arrow space straight s space equals space straight g over 18 space straight x space fraction numerator 2 straight h over denominator straight g end fraction space space space space open parentheses therefore space equals space square root of fraction numerator 2 straight h over denominator straight g end fraction end root close parentheses

therefore s = h/9 m
Hence, the position of ball from the ground= h- h/9 = 8h/9 m

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