A ball is dropped from the top of a tower. In the last second of motion it covers `(5)/(9)` th of tower height before hitting the ground. Acceleration due to gravity = 10 (in `ms^(–2))`. Find height of tower (in m):

Explanation

To solve the problem, we need to find the height of the tower from which a ball is dropped, given that it covers \( \frac{5}{9} \) of the tower height in the last second of its fall. The acceleration due to gravity \( g \) is given as \( 10 \, \text{m/s}^2 \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- Let the height of the tower be \( h \).
- The ball is dropped, so the initial velocity \( u = 0 \).
- The ball covers \( \frac{5}{9}h \) in the last second before hitting the ground.
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