Let A, B and C be points in a vertical line at height h, `(4h)/(5) and (h)/(5) ` and from the ground. A particle released from rest from the point A reaches the ground in time T. The time taken by the particle to go from B to C is ` T_(BC) ` . Then ` (T)/(T_(BC)) ^(2) = `

Explanation

To solve the problem, we need to find the ratio \( \frac{T}{(T_{BC})^2} \) where \( T \) is the time taken for a particle to fall from point A to the ground and \( T_{BC} \) is the time taken for the particle to fall from point B to point C.

### Step-by-Step Solution:

1. **Identify Heights**:
- Height of point A from the ground: \( h \)
- Height of point B from the ground: \( \frac{4h}{5} \)
- Height of point C from the ground: \( \frac{h}{5} \)
...