A body is projected with an initial Velocity 20 m/s at 60° to the horizontal. Its velocity after 1 sec is

Options

A. `10i + 7.32 j`
B. `10i - 7.32j`
C. `10i + sqrt(3) j`
D. `10i-sqrt(3)j`

Answer

To find the velocity of a body projected with an initial velocity of 20 m/s at an angle of 60° to the horizontal after 1 second, we can break the problem down into several steps.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- Initial velocity \( u = 20 \, \text{m/s} \)
- Angle of projection \( \theta = 60^\circ \)

2. **Resolve the Initial Velocity into Components:**
- Horizontal component of the initial velocity:
\[ u_h = u \cos \theta = 20 \cos 60^\circ \]
Since \( \cos 60^\circ = \frac{1}{2} \):
\[ u_h = 20 \times \frac{1}{2} = 10 \, \text{m/s} \]

- Vertical component of the initial velocity:
\[ u_v = u \sin \theta = 20 \sin 60^\circ \]
Since \( \sin 60^\circ = \frac{\sqrt{3}}{2} \):
\[ u_v = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{m/s} \]

3. **Calculate the Horizontal Velocity after 1 Second:**
- There is no horizontal acceleration, so the horizontal velocity remains constant:
\[ V_h = u_h = 10 \, \text{m/s} \]

4. **Calculate the Vertical Velocity after 1 Second:**
- The vertical velocity changes due to gravity. The acceleration due to gravity \( g \) is approximately \( 10 \, \text{m/s}^2 \) (acting downwards).
- The vertical velocity after 1 second is given by:
\[ V_v = u_v - g \cdot t = 10\sqrt{3} - 10 \cdot 1 \]
\[ V_v = 10\sqrt{3} - 10 \]

5. **Combine the Components to Get the Resultant Velocity Vector:**
- The velocity vector after 1 second can be expressed as:
\[ \mathbf{V} = V_h \hat{i} + V_v \hat{j} \]
Substituting the values:
\[ \mathbf{V} = 10 \hat{i} + (10\sqrt{3} - 10) \hat{j} \]

6. **Final Result:**
- Thus, the velocity after 1 second is:
\[ \mathbf{V} = 10 \hat{i} + (10\sqrt{3} - 10) \hat{j} \, \text{m/s} \]