STATEMENT - 1 : Two bodies projected vertically upward with different initial speeds cover same distances in last second of their upward journey.
and
STATEMENT - 2 : For uniformly accelerated motion, distance travelled in a time interval before coming to rest does not depend on initial speed.

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1
B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1
C. Statement-1 is True, Statement-2 is False
D. Statement-1 is False, Statement-2 is True

Answer

To solve the question, we will analyze both statements step by step.

### Step 1: Understanding Statement 1
**Statement 1:** Two bodies projected vertically upward with different initial speeds cover the same distances in the last second of their upward journey.

1. **Let’s denote the two bodies as Body A and Body B.**
- Body A is projected with an initial speed \( u_1 \).
- Body B is projected with an initial speed \( u_2 \).

2. **Both bodies reach their maximum height where their final velocity \( v = 0 \).**
- The time taken to reach maximum height for each body can be calculated using the equation:
\[ v = u - gt \]
- Rearranging gives:
\[ t = \frac{u}{g} \]
- Thus, the time taken for Body A is \( t_1 = \frac{u_1}{g} \) and for Body B is \( t_2 = \frac{u_2}{g} \).

3. **Distance covered in the last second of their upward journey:**
- The distance covered in the last second for any uniformly accelerated motion can be calculated using:
\[ S_n = u t + \frac{1}{2} a t^2 \]
- For the last second, we take \( t = 1 \) second before reaching the maximum height.
- For Body A:
\[ S_{nA} = u_1(1) - \frac{1}{2}g(1^2) = u_1 - \frac{g}{2} \]
- For Body B:
\[ S_{nB} = u_2(1) - \frac{1}{2}g(1^2) = u_2 - \frac{g}{2} \]

4. **Setting the distances equal:**
- Since both bodies cover the same distance in the last second:
\[ u_1 - \frac{g}{2} = u_2 - \frac{g}{2} \]
- This implies:
\[ u_1 = u_2 \]
- Therefore, the statement is true.

### Step 2: Understanding Statement 2
**Statement 2:** For uniformly accelerated motion, the distance traveled in a time interval before coming to rest does not depend on initial speed.

1. **Consider an object projected upwards until it comes to rest.**
- The distance covered before coming to rest can be calculated similarly using:
\[ S = ut + \frac{1}{2} a t^2 \]
- Here, \( u \) is the initial speed, \( a = -g \) (acceleration due to gravity), and the time until it comes to rest is \( t = \frac{u}{g} \).

2. **Substituting into the distance formula:**
- The total distance \( S \) before coming to rest is:
\[ S = u \left(\frac{u}{g}\right) + \frac{1}{2}(-g)\left(\frac{u}{g}\right)^2 \]
- Simplifying gives:
\[ S = \frac{u^2}{g} - \frac{1}{2} \frac{u^2}{g} = \frac{u^2}{2g} \]
- This shows that the distance does depend on the initial speed \( u \).

### Conclusion
Both statements are true, but the second statement is not a correct explanation for the first statement.

### Final Answer
- **Statement 1 is true.**
- **Statement 2 is true.**
- **Statement 2 is not the correct explanation for Statement 1.**