An object is projected with velocity `vecv_(0) = 15hati + 20hatj` . Considering x along horizontal axis and y along vertical axis. Find its velocity after 2s.

Answer

To find the velocity of the object after 2 seconds, we can use the kinematic equation for velocity:

### Step-by-Step Solution:

1. **Identify the Initial Velocity Vector**:
The initial velocity vector \( \vec{v_0} \) is given as:
\[ \vec{v_0} = 15 \hat{i} + 20 \hat{j} \quad \text{(in m/s)} \]

2. **Determine the Acceleration**:
Since the object is projected in a gravitational field, the only acceleration acting on it is due to gravity. We will take the acceleration due to gravity \( g \) as:
\[ \vec{a} = -g \hat{j} = -10 \hat{j} \quad \text{(in m/s}^2\text{)} \]
The negative sign indicates that gravity acts downwards.

3. **Use the Kinematic Equation for Velocity**:
The formula for final velocity \( \vec{v} \) is given by:
\[ \vec{v} = \vec{v_0} + \vec{a} \cdot t \]
where \( t \) is the time interval. Here, \( t = 2 \) seconds.

4. **Substitute the Values**:
Now we substitute the values into the equation:
\[ \vec{v} = (15 \hat{i} + 20 \hat{j}) + (-10 \hat{j}) \cdot 2 \]

5. **Calculate the Acceleration Component**:
Calculate the acceleration component:
\[ \vec{a} \cdot t = (-10 \hat{j}) \cdot 2 = -20 \hat{j} \]

6. **Combine the Vectors**:
Now substitute this back into the velocity equation:
\[ \vec{v} = 15 \hat{i} + 20 \hat{j} - 20 \hat{j} \]

7. **Simplify the Expression**:
Combine the \( \hat{j} \) components:
\[ \vec{v} = 15 \hat{i} + (20 - 20) \hat{j} = 15 \hat{i} + 0 \hat{j} \]

8. **Final Result**:
Thus, the final velocity after 2 seconds is:
\[ \vec{v} = 15 \hat{i} \quad \text{(in m/s)} \]

### Conclusion:
The velocity of the object after 2 seconds is \( 15 \hat{i} \) m/s, indicating that it has only horizontal velocity and no vertical velocity.