A body is allowed to fall from a height of `10 m`. If the time taken for the first `50 m` is `t_(1)` and for the remaining `50 s`,is `t_(2)`.
The ratio `t_(1)` and `t_(2)`. Is nearly .

Options

A. `5:2`
B. `3:1`
C. `3:2`
D. `5: 3`

Explanation

To solve the problem, we will use the equations of motion under uniform acceleration due to gravity. We are given that a body is dropped from a height of 100 m, and we need to find the time taken to fall the first 50 m (denoted as \( t_1 \)) and the time taken to fall the remaining 50 m (denoted as \( t_2 \)). Finally, we will calculate the ratio \( \frac{t_1}{t_2} \).

### Step-by-Step Solution:

1. **Calculate \( t_1 \) for the first 50 m:**
We can use the second equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
Here, \( s = 50 \, \text{m} \), \( u = 0 \, \text{m/s} \) (initial velocity), and \( a = g = 9.8 \, \text{m/s}^2 \). The equation simplifies to:
\[ 50 = 0 + \frac{1}{2} g t_1^2 \]
Rearranging gives:
\[ t_1^2 = \frac{100}{g} \]
Therefore:
\[ t_1 = \sqrt{\frac{100}{g}} = \sqrt{\frac{100}{9.8}} \approx \sqrt{10.2} \approx 3.2 \, \text{s} \]

2. **Calculate the total time \( t \) to fall 100 m:**
Using the same equation for the total distance:
\[ s = 100 \, \text{m} \]
\[ 100 = 0 + \frac{1}{2} g t^2 \]
Rearranging gives:
\[ t^2 = \frac{200}{g} \]
Therefore:
\[ t = \sqrt{\frac{200}{g}} = \sqrt{\frac{200}{9.8}} \approx \sqrt{20.4} \approx 4.5 \, \text{s} \]

3. **Calculate \( t_2 \) for the remaining 50 m:**
The time taken to fall the remaining 50 m is:
\[ t_2 = t - t_1 \]
Substituting the values we found:
\[ t_2 = 4.5 - 3.2 \approx 1.3 \, \text{s} \]

4. **Calculate the ratio \( \frac{t_1}{t_2} \):**
Now we can find the ratio:
\[ \frac{t_1}{t_2} = \frac{3.2}{1.3} \approx 2.46 \]
This can be approximated to a ratio of \( \frac{5}{2} \) when expressed in simplest form.

### Final Answer:
The ratio \( \frac{t_1}{t_2} \) is nearly \( 5:2 \).