An object is projected with velocity `vecv_(0) = 15hatj + 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 will analyze the motion in both the x and y directions separately.

### Step 1: Identify the initial velocity components
The initial velocity vector is given as:
\[ \vec{v_0} = 15 \hat{i} + 20 \hat{j} \]
From this, we can identify the components of the initial velocity:
- \(v_{0x} = 15 \, \text{m/s}\) (horizontal component)
- \(v_{0y} = 20 \, \text{m/s}\) (vertical component)

### Step 2: Analyze the motion in the x direction
In the x direction, there is no acceleration acting on the object (assuming no air resistance). Therefore, the horizontal velocity remains constant:
\[ v_x = v_{0x} = 15 \, \text{m/s} \]

### Step 3: Analyze the motion in the y direction
In the y direction, the object is subject to gravitational acceleration, which acts downwards. The acceleration due to gravity is:
\[ g = 10 \, \text{m/s}^2 \]
The velocity in the y direction after time \(t\) can be calculated using the equation:
\[ v_y = v_{0y} - g \cdot t \]
Substituting the known values:
\[ v_y = 20 \, \text{m/s} - (10 \, \text{m/s}^2 \cdot 2 \, \text{s}) = 20 - 20 = 0 \, \text{m/s} \]

### Step 4: Combine the velocity components
After 2 seconds, the velocity components are:
- \(v_x = 15 \, \text{m/s}\)
- \(v_y = 0 \, \text{m/s}\)

Thus, the resultant velocity vector after 2 seconds is:
\[ \vec{v} = 15 \hat{i} + 0 \hat{j} = 15 \hat{i} \]

### Final Answer
The velocity of the object after 2 seconds is:
\[ \vec{v} = 15 \hat{i} \, \text{m/s} \]

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