A body is dropped from rest from a height h. It covers a distance `(9h)/(25)` in the last seconds of fall. The height h is : (use `g = 9.8 m//sec^(2)` )

Options

A. `102.5 m`
B. `112.5 m`
C. `122.5 m`
D. `132.5 m`

Explanation

To solve the problem, we need to determine the height \( h \) from which a body is dropped, given that it covers a distance of \( \frac{9h}{25} \) in the last second of its fall.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A body is dropped from a height \( h \).
- It covers a distance of \( \frac{9h}{25} \) in the last second of its fall.

2. **Using the Equations of Motion**:
- The distance covered in free fall under gravity can be described using the equation:
\[ h = ut + \frac{1}{2} g t^2 \]
- Since the body is dropped from rest, the initial velocity \( u = 0 \). Thus, the equation simplifies to:
\[ h = \frac{1}{2} g t^2 \]

3. **Distance Covered in the Last Second**:
- The distance covered in the last second can be expressed as:
\[ d = h - h_{t-1} \]
- Where \( h_{t-1} \) is the distance covered in the first \( t-1 \) seconds:
\[ h_{t-1} = \frac{1}{2} g (t-1)^2 \]
- Therefore, the distance covered in the last second is:
\[ d = h - h_{t-1} = h - \frac{1}{2} g (t-1)^2 \]

4. **Setting Up the Equation**:
- We know that in the last second, the distance covered is \( \frac{9h}{25} \):
\[ \frac{9h}{25} = h - \frac{1}{2} g (t-1)^2 \]
- Substitute \( h = \frac{1}{2} g t^2 \) into the equation:
\[ \frac{9}{25} \left(\frac{1}{2} g t^2\right) = \frac{1}{2} g t^2 - \frac{1}{2} g (t-1)^2 \]

5. **Simplifying the Equation**:
- Multiply through by \( 2 \) to eliminate the fraction:
\[ \frac{9}{25} g t^2 = g t^2 - g (t-1)^2 \]
- Dividing by \( g \) (assuming \( g \neq 0 \)):
\[ \frac{9}{25} t^2 = t^2 - (t^2 - 2t + 1) \]
- Simplifying further:
\[ \frac{9}{25} t^2 = 2t - 1 \]

6. **Rearranging the Equation**:
- Rearranging gives:
\[ \frac{9}{25} t^2 - 2t + 1 = 0 \]
- Multiply through by \( 25 \) to clear the fraction:
\[ 9t^2 - 50t + 25 = 0 \]

7. **Using the Quadratic Formula**:
- The quadratic formula is:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- Here, \( a = 9, b = -50, c = 25 \):
\[ t = \frac{50 \pm \sqrt{(-50)^2 - 4 \cdot 9 \cdot 25}}{2 \cdot 9} \]
- Calculate the discriminant:
\[ t = \frac{50 \pm \sqrt{2500 - 900}}{18} = \frac{50 \pm \sqrt{1600}}{18} = \frac{50 \pm 40}{18} \]
- This gives two possible values for \( t \):
\[ t = \frac{90}{18} = 5 \quad \text{or} \quad t = \frac{10}{18} = \frac{5}{9} \]

8. **Finding the Height \( h \)**:
- We take \( t = 5 \) seconds (as time cannot be negative or less than 1 second).
- Substitute \( t \) back into the height equation:
\[ h = \frac{1}{2} g t^2 = \frac{1}{2} \cdot 9.8 \cdot 5^2 = \frac{1}{2} \cdot 9.8 \cdot 25 = 122.5 \, \text{meters} \]

### Final Answer:
The height \( h \) is \( 122.5 \) meters.