A ball is released from the top of a building of height `45m` at `t=0` . When the ball strikes the ground, its speed becomes `(2)/(3)` times its original value. The time instant when the ball strikes the ground for the second time is t = ...........sec .`["Take" g=10 m//s^(2) ]`

Explanation

To solve the problem step by step, we will break down the motion of the ball into two parts: the first fall to the ground and the second rise and fall back to the ground.

### Step 1: Calculate the time taken for the first fall (t1)

The ball is dropped from a height of 45 m. We can use the equation of motion:

\[
s = ut + \frac{1}{2} a t^2
...