A body is projected horizontally with a velocity of `4 ms^(-1)` from the top of a high tower. The velocity of the body after 0.7 is nearly (Take, `g = 10 ms^(-2))`

Options

A. `10 ms^(-1)`
B. `8 ms^(-1)`
C. `19.2 ms^(-1)`
D. `11 ms^(-1)`

Explanation

To find the velocity of the body after 0.7 seconds, we can break the problem down into horizontal and vertical components.

### Step-by-Step Solution:

1. **Identify the Initial Conditions**:
- The body is projected horizontally with an initial horizontal velocity, \( V_x = 4 \, \text{m/s} \).
- The initial vertical velocity, \( V_y = 0 \, \text{m/s} \) (since it is projected horizontally).
- The acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \).

2. **Calculate the Vertical Velocity after 0.7 seconds**:
- The vertical velocity can be calculated using the formula:
\[ V_y = g \cdot t \]
- Substituting the values:
\[ V_y = 10 \, \text{m/s}^2 \cdot 0.7 \, \text{s} = 7 \, \text{m/s} \]
- This velocity is directed downward.

3. **Determine the Horizontal Velocity**:
- The horizontal velocity remains constant since there is no horizontal acceleration:
\[ V_x = 4 \, \text{m/s} \]

4. **Calculate the Resultant Velocity**:
- The resultant velocity \( V \) can be found using the Pythagorean theorem:
\[ V = \sqrt{V_x^2 + V_y^2} \]
- Substituting the values:
\[ V = \sqrt{(4 \, \text{m/s})^2 + (7 \, \text{m/s})^2} = \sqrt{16 + 49} = \sqrt{65} \, \text{m/s} \]

5. **Approximate the Resultant Velocity**:
- Since \( \sqrt{65} \) is close to \( \sqrt{64} \), we can approximate:
\[ \sqrt{65} \approx 8 \, \text{m/s} \]

6. **Final Answer**:
- The velocity of the body after 0.7 seconds is approximately \( 8 \, \text{m/s} \).