Introduction:
In this problem, we are given that a ball is thrown vertically upwards from the ground and it crosses a point at a height of 25m twice at an interval of 4s. We need to determine the initial velocity with which the ball was thrown.

Key Information:
- The ball is thrown vertically upwards, which means its motion is only in the vertical direction.
- The ball crosses a point at a height of 25m twice at an interval of 4s.

Understanding the Problem:
To solve this problem, we need to understand the motion of the ball and use the given information to find its initial velocity.

Vertical Motion of the Ball:
When the ball is thrown upwards, it experiences a vertical motion under the influence of gravity. The key parameters involved in the motion are:
- Initial velocity (u): The velocity with which the ball is thrown upwards.
- Final velocity (v): The velocity of the ball at any given point in its motion.
- Acceleration due to gravity (g): The acceleration acting on the ball in the vertical direction.

Using the Equations of Motion:
We can use the equations of motion to solve this problem. The relevant equation for the vertical motion of the ball is:
v = u + gt

First Crossing of the 25m Point:
Let's consider the first crossing of the 25m point. At this point, the displacement of the ball is 25m and the time taken to reach this point is 4s.
Using the equation of motion, we can write:
25 = u * 4 + (0.5) * (-9.8) * (4)^2

Simplifying the equation, we get:
25 = 4u - 78.4

Solving for u, we find:
4u = 103.4
u = 25.85 m/s

Second Crossing of the 25m Point:
Now, let's consider the second crossing of the 25m point. At this point, the displacement of the ball is again 25m, but the time taken to reach this point is given as an interval of 4s.
This means the total time taken for the ball to reach the second crossing point is 4s + 4s = 8s.

Using the equation of motion, we can write:
25 = u * 8 + (0.5) * (-9.8) * (8)^2

Simplifying the equation, we get:
25 = 8u - 313.6

Solving for u, we find:
8u = 338.6
u = 42.325 m/s

Conclusion:
The initial velocity with which the ball was thrown vertically upwards is approximately 42.325 m/s.

30 m/s
as from tower ball will be in air for 2 sec till max height so max height will be 45 m so by formula h = u square / 2g ; u will be 30 m/s