A ball of mass 200g is projected with a density of `30m//s` at `30^@` from horizontal. Using the concept of impulse, find change in velocity in 2s. Take `g=10m//s^2`.

Explanation

To solve the problem, we will use the concept of impulse and the relationship between impulse and momentum. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Given Data
- Mass of the ball, \( m = 200 \, \text{g} = 0.2 \, \text{kg} \) (since 1 g = 0.001 kg)
- Initial velocity, \( v_0 = 30 \, \text{m/s} \)
- Angle of projection, \( \theta = 30^\circ \)
- Time duration, \( t = 2 \, \text{s} \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the Weight of the Ball
The weight \( W \) of the ball can be calculated using the formula:
\[ W = m \cdot g \]
Substituting the values:
\[ W = 0.2 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 2 \, \text{N} \]

### Step 3: Calculate the Impulse
Impulse \( J \) is defined as the product of force and time:
\[ J = F \cdot t \]
In this case, the force \( F \) acting on the ball is its weight:
\[ J = W \cdot t = 2 \, \text{N} \cdot 2 \, \text{s} = 4 \, \text{Ns} \]

### Step 4: Relate Impulse to Change in Momentum
Impulse is also equal to the change in momentum:
\[ J = \Delta p \]
Where \( \Delta p = m \cdot \Delta v \) and \( \Delta v \) is the change in velocity. Therefore:
\[ \Delta p = J \Rightarrow m \cdot \Delta v = 4 \, \text{Ns} \]

### Step 5: Solve for Change in Velocity
Rearranging the equation to find \( \Delta v \):
\[ \Delta v = \frac{J}{m} = \frac{4 \, \text{Ns}}{0.2 \, \text{kg}} = 20 \, \text{m/s} \]

### Conclusion
The change in velocity of the ball after 2 seconds is \( 20 \, \text{m/s} \).

---