A ball is dropped from the top of a tower of height (h). It covers a distance of `h`/`2` in the last second of its motion. How long does the ball remain in air?

Explanation

Let the ball dropped grom the top the tower of height (h) taken time (t) while reaching the ground. Using the relation for the distance travelled in ltBrgt ` nth second, ` D_n =u + a/2 ( 2 n-1) , we have
h/2 = 0 + a/2 ( 2 xx t t-1) ` …(i)
Using`, ` S= ut + 1/2 + 1/2 at^2`, we have ltBrgt ` h=0 + 1/2 at^2` ..(ii)
Solving (i) and (ii) we get ` t= ( 2 +- sqrt 2) s`
Max, time for which the ball remains in air
`= ( 2 + sqrt 2) s` .