A body is projected horizontally from a point above the ground and motion of the body is described by the equation x=2t, `y=5t^2` where x, and y are horizontal and vertical coordinates in metre after time t. The initial velocity of the body will be

Options

A. `sqrt29` m/s horizontal
B. 5m/s horizontal
C. 2m/s vertical
D. 2 m/s horizontal

Explanation

To find the initial velocity of the body projected horizontally, we can analyze the given equations of motion:

1. **Given Equations**:
- Horizontal motion: \( x = 2t \)
- Vertical motion: \( y = 5t^2 \)

2. **Understanding the Equations**:
- The equation for horizontal motion \( x = 2t \) indicates that the horizontal position \( x \) changes linearly with time \( t \). The coefficient of \( t \) (which is 2) represents the horizontal velocity.
- The equation for vertical motion \( y = 5t^2 \) indicates that the vertical position \( y \) changes with the square of time \( t \), which is characteristic of uniformly accelerated motion (in this case, due to gravity).

3. **Finding the Initial Velocity**:
- The horizontal velocity \( v_x \) can be found directly from the equation \( x = 2t \). The horizontal velocity is the coefficient of \( t \) in the equation:
\[ v_x = \frac{dx}{dt} = 2 \, \text{m/s} \]

4. **Conclusion**:
- Since the body is projected horizontally, the initial velocity of the body is equal to the horizontal velocity:
\[ \text{Initial Velocity} = 2 \, \text{m/s} \]

### Summary:
The initial velocity of the body projected horizontally is \( 2 \, \text{m/s} \).