When a particle is undergoing motion, the diplacement of the particle has a magnitude that is equal to or smaller than the total distance travelled by the particle. In many cases the displacement of the particle may actually be zero, while the distance travelled by it is non-zero. Both these quantities, however depend on the frame of reference in which motion of the particle is being observed. Consider a particle which is projected in the earth's gravitational field, close to its surface, with a speed of `100sqrt(2) m//s`, at an angle of `45^(@)` with the horizontal in the eastward direction. Ignore air resistance and assume that the acceleration due to gravity is `10 m//s^(2)`.
" A third observer (C) close to the surface of the reports that particle is initially travelling at a speed of `100sqrt(2) m//s` making on angle of `45^(@)` with the horizontal, but its horizontal motion is northward". The third observer is moving in :-

Options

A. The south-west direction with a speed of `100 sqrt(2) m//s`
B. The south-east direction with a speed of `100 sqrt(2) m//s`
C. The north-west direction with a speed of `100 sqrt(2) m//s`
D. The north-east direction with speed of `100 sqrt(2) m//s`

Explanation

To solve the problem, we need to analyze the motion of the particle and the observer's frame of reference. Let's break it down step by step:

### Step 1: Understand the Initial Conditions
The particle is projected with a speed of \(100\sqrt{2} \, \text{m/s}\) at an angle of \(45^\circ\) with the horizontal in the eastward direction. This means the initial velocity components can be calculated as follows:
- Horizontal (Eastward) component:
\[ v_{ox} = 100\sqrt{2} \cdot \cos(45^\circ) = 100\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 100 \, \text{m/s} \]
- Vertical component:
\[ v_{oy} = 100\sqrt{2} \cdot \sin(45^\circ) = 100\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 100 \, \text{m/s} \]

### Step 2: Analyze the Observer's Frame of Reference
The observer (C) reports that the particle is moving with the same speed of \(100\sqrt{2} \, \text{m/s}\) but in a northward direction. This implies that the observer must be moving in such a way that the relative motion of the particle appears northward to them.

### Step 3: Determine the Observer's Velocity
To find the observer's velocity, we denote it as \(v_c\). The observer's velocity must counteract the eastward motion of the particle so that the resultant motion appears to be northward.

1. **Assuming the observer moves in the southwest direction:**
- If the observer moves southwest at \(100\sqrt{2} \, \text{m/s}\), the components of the observer's velocity would be:
\[ v_{cx} = -100 \, \text{m/s} \quad (\text{Westward}) \]
\[ v_{cy} = -100 \, \text{m/s} \quad (\text{Southward}) \]
- The relative velocity of the particle with respect to the observer would be:
\[ v_{co} = v_{ox} - v_{cx} = 100 - (-100) = 200 \, \text{m/s} \quad (\text{Eastward}) \]
- This does not result in a northward motion, so this option is incorrect.

2. **Assuming the observer moves in the southeast direction:**
- The components would be:
\[ v_{cx} = 100 \, \text{m/s} \quad (\text{Eastward}) \]
\[ v_{cy} = -100 \, \text{m/s} \quad (\text{Southward}) \]
- The relative velocity would be:
\[ v_{co} = v_{ox} - v_{cx} = 100 - 100 = 0 \quad (\text{Eastward}) \]
- This does not yield a northward motion, so this option is incorrect.

3. **Assuming the observer moves in the northwest direction:**
- The components would be:
\[ v_{cx} = -100 \, \text{m/s} \quad (\text{Westward}) \]
\[ v_{cy} = 100 \, \text{m/s} \quad (\text{Northward}) \]
- The relative velocity would be:
\[ v_{co} = v_{ox} - v_{cx} = 100 - (-100) = 200 \, \text{m/s} \quad (\text{Eastward}) \]
- This does not yield a northward motion, so this option is incorrect.

4. **Assuming the observer moves in the northeast direction:**
- The components would be:
\[ v_{cx} = 100 \, \text{m/s} \quad (\text{Eastward}) \]
\[ v_{cy} = 100 \, \text{m/s} \quad (\text{Northward}) \]
- The relative velocity would be:
\[ v_{co} = v_{ox} - v_{cx} = 100 - 100 = 0 \quad (\text{Eastward}) \]
- This does not yield a northward motion, so this option is incorrect.

### Conclusion
From the analysis, the only viable option that allows the observer to see the particle moving northward is if the observer is moving in the southeast direction at a speed of \(100\sqrt{2} \, \text{m/s}\).

### Final Answer
The third observer (C) is moving in the southeast direction with a speed of \(100\sqrt{2} \, \text{m/s}\).