A particle is droped under gravety from rest a height h and it travels a distance `(9h)/(25)`in the last second, the height h is (take `g=9.8ms^(-2)`)

Explanation

To solve the problem, we need to find the height \( h \) from which a particle is dropped under gravity, given that it travels a distance of \( \frac{9h}{25} \) in the last second of its fall. We will use the equations of motion under uniform acceleration due to gravity.

### Step-by-Step Solution:

1. **Understanding the Motion**:
The particle is dropped from rest, so its initial velocity \( u = 0 \). The distance traveled in the \( n^{th} \) second is given by the formula:
\[
S_n = u + \frac{1}{2} g (2n - 1)
...