A body is projected horizontally from certain height. After `sqrt(3)` seconds, its direction of motion makes `60^(@)` with the horizontal . Then its initial velocity of projection is

Options

A. `9.8 ms^(-1)`
B. `4.9 ms^(-1)`
C. `19.6 ms^(-1)`
D. `14.7 ms^(-1)`

Answer

To solve the problem, we need to find the initial velocity of a body projected horizontally from a certain height, given that after \(\sqrt{3}\) seconds, its direction of motion makes an angle of \(60^\circ\) with the horizontal.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The body is projected horizontally, meaning its initial vertical velocity is \(0\).
- The horizontal velocity remains constant throughout the motion because there is no horizontal acceleration.

2. **Velocity Components After \(\sqrt{3}\) Seconds**:
- Let the initial horizontal velocity be \(v\).
- After \(\sqrt{3}\) seconds, the velocity of the body can be broken down into horizontal and vertical components.
- The horizontal component of the velocity after \(\sqrt{3}\) seconds remains \(v\).
- The vertical component of the velocity after \(\sqrt{3}\) seconds can be calculated using the formula for vertical motion under gravity:
\[ v_y = u + at \]
where \(u = 0\) (initial vertical velocity), \(a = g = 9.8 \, \text{m/s}^2\) (acceleration due to gravity), and \(t = \sqrt{3}\) seconds.

3. **Calculating the Vertical Velocity**:
- Plugging in the values:
\[ v_y = 0 + 9.8 \cdot \sqrt{3} \]
\[ v_y = 9.8\sqrt{3} \]

4. **Using the Angle of Motion**:
- The direction of motion makes an angle of \(60^\circ\) with the horizontal.
- The relationship between the components of the velocity and the angle is given by:
\[ \tan(60^\circ) = \frac{v_y}{v} \]
- Since \(\tan(60^\circ) = \sqrt{3}\), we can write:
\[ \sqrt{3} = \frac{9.8\sqrt{3}}{v} \]

5. **Solving for Initial Velocity \(v\)**:
- Rearranging the equation gives:
\[ v = 9.8 \]

### Final Answer:
The initial velocity of projection is \(9.8 \, \text{m/s}\).