Question

A freely falling body takes t' second to travel first (1/x)th distance. Then, time of descent is​

Answer 1

Answer:

Let total distance be y

So, y(1/x) = 1/2gt²

y = (x/2)gt²

Now, time of descent, T= $\sqrt{2y/g}$

T = $\sqrt{(2/g)*(x/2)gt^{2} }$

T = t√x