A ball is dropped from the roof of a tower of height (h). The total distanc coverd by it in the last second of its motion is equal to the distance covered by in first three seconds. What will be the velocity at the end of `9 secind ` ?

Explanation

Let the ball remain in air for (t) seconds. Then ,
` D_1 = u + a/ 2 ( 2t- 1) =0 + ( 10) /2 ( 2t-t)`
` = 10 t-5` …(i)
Distance covered in first three secnds.
` S= 1/2 gt^2 = 1/2 xx 10 3^2 = 45` …(ii)
As per question ` 10 t- 5 = 45`
or ` t= 5 s`
:. ` h = 1/2 at^2 = 1/2 xx 10 xx 10 5^2 = 125 m` .