A freely falling body takes `2` seconds to reach the ground on a planet when it is dropped from a height of `8m`. If the period of a simple pondulum is `2` seconds on that planet then its length is

Options

A. `40cm`
B. `50cm`
C. `75cm`
D. `25cm`

Explanation

To solve the problem, we need to find the length of a simple pendulum on a planet where a freely falling body takes 2 seconds to fall from a height of 8 meters. We also know that the period of the pendulum is 2 seconds.

### Step-by-Step Solution:

1. **Determine the acceleration due to gravity (g) on the planet:**
We can use the equation of motion for a freely falling body:
\[ S = ut + \frac{1}{2} a t^2 \]
Here, \( S = 8 \, \text{m} \), \( u = 0 \, \text{m/s} \) (initial velocity), and \( t = 2 \, \text{s} \). Plugging in these values:
\[ 8 = 0 \cdot 2 + \frac{1}{2} a (2^2) \]
Simplifying this gives:
\[ 8 = \frac{1}{2} a \cdot 4 \]
\[ 8 = 2a \]
\[ a = \frac{8}{2} = 4 \, \text{m/s}^2 \]

2. **Use the formula for the period of a simple pendulum:**
The formula for the period \( T \) of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
We know that \( T = 2 \, \text{s} \) and \( g = 4 \, \text{m/s}^2 \). Plugging in these values:
\[ 2 = 2\pi \sqrt{\frac{L}{4}} \]

3. **Simplify the equation:**
Dividing both sides by 2:
\[ 1 = \pi \sqrt{\frac{L}{4}} \]
Now, squaring both sides:
\[ 1^2 = \pi^2 \left(\frac{L}{4}\right) \]
\[ 1 = \frac{\pi^2 L}{4} \]

4. **Solve for L:**
Rearranging the equation gives:
\[ L = \frac{4}{\pi^2} \]

5. **Calculate the numerical value of L:**
Using \( \pi \approx 3.14 \):
\[ L = \frac{4}{(3.14)^2} \approx \frac{4}{9.8596} \approx 0.405 \, \text{m} \]

6. **Convert L to centimeters:**
To convert meters to centimeters, we multiply by 100:
\[ L \approx 0.405 \times 100 \approx 40.5 \, \text{cm} \]

### Final Answer:
The length of the pendulum is approximately **40.5 cm**.