A ball is thrown vertically upward from the ground. It crosses a point at the height of 25 m twice at an interval of 4 seconds. The ball was thrown with the velocity of

The interval between object pass the same point is 4 sec.

That means in 2 sec, object reaches the top and in next 2 sec, it again reaches the same point.

By the info given, we can find the velocity of that point by using : v=u+(-gt)

0= u - 10×2

u = 20

Then the velocity at 25m is 20m/sec.

so the initial velocity is

$$\begin{gathered} - v 2 = u 2 + ( - 2 gh ) \\ 400 = u 2 - 2 x 10 x 25 \\ u 2 = 900 \\ = > u = 30 m / s \end{gathered}$$