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A ball is released from the top of a tow...

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

A

`h//9` meter from the ground

B

`7h//9` meter from the ground

C

`8h//9` meter from the ground

D

`17h//18` meter from the ground

Text Solution

Verified by Experts

The correct Answer is:
C
Second law of motion gives
`s = ut +(1)/(2) "gt"^(2)`
or `h = 0 +(1)/(2) "g t"^(2)` (`:' U = 0)`
`:.T = sqrt(((2h)/(g)))`
At `t = (T)/(3) sec`,
`s = 0 +(1)/(2)g ((T)/(3))^(2)`
`rArr s = (1)/(2)g. (T^(2))/(9)`
`rArr s = (g)/(18) xx (2h)/(g) (:' T = sqrt((2h)/(g)))`
`:. s = (h)/(9) m`
Hence, the position of ball from teh ground
`= h - (h)/(9) = (8h)/(9) m`
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