A body freely falling from a height h describes `(11h)/36` in the last second of its fall. The height h is `(g = 10 ms^(-2))`

Options

A. 125 m
B. 180 m
C. 360 m
D. 90 m

Answer

To solve the problem, we need to find the height \( h \) from which a body is freely falling, given that it covers \( \frac{11h}{36} \) in the last second of its fall, with \( g = 10 \, \text{m/s}^2 \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The body falls from a height \( h \) and takes \( t \) seconds to reach the ground.
- In the last second of its fall (from \( t-1 \) to \( t \)), it covers a distance of \( \frac{11h}{36} \).

2. **Distance Covered in Total Time**:
- The total distance \( h \) fallen in \( t \) seconds can be expressed using the equation of motion:
\[ h = \frac{1}{2} g t^2 \]
- Substituting \( g = 10 \, \text{m/s}^2 \):
\[ h = 5t^2 \quad \text{(Equation 1)} \]

3. **Distance Covered in the Last Second**:
- The distance covered in the last second (from \( t-1 \) to \( t \)) can also be expressed as:
\[ \text{Distance in last second} = h - \text{Distance covered in } (t-1) \text{ seconds} \]
- The distance covered in \( t-1 \) seconds is:
\[ \text{Distance} = \frac{1}{2} g (t-1)^2 = 5(t-1)^2 \]
- Therefore, the distance covered in the last second is:
\[ \frac{11h}{36} = h - 5(t-1)^2 \]

4. **Substituting \( h \) from Equation 1**:
- Substitute \( h = 5t^2 \) into the distance equation:
\[ \frac{11(5t^2)}{36} = 5t^2 - 5(t-1)^2 \]
- Simplifying gives:
\[ \frac{55t^2}{36} = 5t^2 - 5(t^2 - 2t + 1) \]
- This simplifies to:
\[ \frac{55t^2}{36} = 5t^2 - 5t^2 + 10t - 5 \]
- Thus:
\[ \frac{55t^2}{36} = 10t - 5 \]

5. **Clearing the Fraction**:
- Multiply through by 36 to eliminate the fraction:
\[ 55t^2 = 360t - 180 \]
- Rearranging gives:
\[ 55t^2 - 360t + 180 = 0 \]

6. **Using the Quadratic Formula**:
- The quadratic formula is:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- Here, \( a = 55 \), \( b = -360 \), and \( c = 180 \):
\[ t = \frac{360 \pm \sqrt{(-360)^2 - 4 \cdot 55 \cdot 180}}{2 \cdot 55} \]
- Calculate the discriminant:
\[ (-360)^2 - 4 \cdot 55 \cdot 180 = 129600 - 39600 = 90000 \]
- Thus:
\[ t = \frac{360 \pm 300}{110} \]
- This gives two possible values for \( t \):
\[ t = \frac{660}{110} = 6 \quad \text{and} \quad t = \frac{60}{110} = \frac{6}{11} \text{ (not valid as time cannot be negative)} \]

7. **Finding the Height**:
- Substitute \( t = 6 \) back into Equation 1 to find \( h \):
\[ h = 5t^2 = 5 \cdot 6^2 = 5 \cdot 36 = 180 \, \text{m} \]

### Final Answer:
The height \( h \) is \( 180 \, \text{m} \).