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

Answer: Option C

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
−v² = u² + (−2gh)
400 = u² − 2x 10 x 25
u² = 900
=> u = 30 m/s