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A body took 't' sec. to come down from t...

A body took 't' sec. to come down from top of tower.Find the time taken to cover half the height of the tower .

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Let AB be the tower of height .h. m and body taken .t. seconds to come down from A to B as shown in the figure.
Let C be the midpoint and hence AC=`(h)/(2)` m

S=ut+1/2`at^(2)`
Applying the above formula for both the cased ,we get
`h=(1)/(2)"gt"^(2)` .............(1) ,`(h)/(2)=(1)/(2) gt^(2)`....... (2)
Eliminating the unconcerned terms by dividing terms by dividing equation (1) by (2),we get
`((1))/((2))=(h)/(h//2)=((1)/(2)"gt"^(2))/((1)/(2)"gt"^(2))implies2=((t)/(t.))^(2)impliest^(t^(2))=(t^(2))/(2)impliest^(1)=(t)/(sqrt(2))s`
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