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)`.
Consider an observer in frame D (of the previous question), who observes a body of mass 10 kg acelerating in the upward direction at `30 m//s^(2)` (w.r.t. himself). The net force acting on this body, as observed from the ground is :-

Options

A. `400 N` in the upward direction
B. `300 N` in the upward direction
C. `200 N` in the upward direction
D. `500 N` in the upward direction

Explanation

To solve the problem, we need to find the net force acting on a body of mass 10 kg that is accelerating upwards at 30 m/s², as observed from the ground. We will apply Newton's second law of motion to find the required net force.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Body**:
- The body has an upward force \( F_{\text{up}} \).
- The weight of the body acts downward, which is given by \( W = m \cdot g \), where \( m \) is the mass of the body and \( g \) is the acceleration due to gravity.

2. **Calculate the Weight of the Body**:
- Given:
- Mass \( m = 10 \, \text{kg} \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)
- Weight \( W = m \cdot g = 10 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 100 \, \text{N} \) (downward).

3. **Apply Newton's Second Law**:
- According to Newton's second law, the net force \( F_{\text{net}} \) acting on the body can be expressed as:
\[ F_{\text{net}} = F_{\text{up}} - W \]
- Here, \( F_{\text{net}} \) is also equal to \( m \cdot a \), where \( a \) is the acceleration of the body.

4. **Substitute the Known Values**:
- The body is accelerating upwards at \( a = 30 \, \text{m/s}^2 \).
- Therefore, we can write:
\[ F_{\text{net}} = m \cdot a = 10 \, \text{kg} \cdot 30 \, \text{m/s}^2 = 300 \, \text{N} \]

5. **Set Up the Equation**:
- Now, substituting the value of \( F_{\text{net}} \) into the equation:
\[ 300 \, \text{N} = F_{\text{up}} - 100 \, \text{N} \]

6. **Solve for \( F_{\text{up}} \)**:
- Rearranging the equation gives us:
\[ F_{\text{up}} = 300 \, \text{N} + 100 \, \text{N} = 400 \, \text{N} \]

7. **Conclusion**:
- The net force acting on the body, as observed from the ground, is \( 400 \, \text{N} \) in the upward direction.

### Final Answer:
The net force acting on the body is **400 N upwards**.