A body is projected horizontally with a velocity of `4sqrt2m//sec`. The velocity of the body after 0.7 seconds will be bearly (Take g = `10 m/sec^(2)`)

Options

A. 10 m/sec
B. 9 m/sec
C. 19 m/sec
D. 11 m/sec

Explanation

To solve the problem step by step, we will analyze the motion of the body projected horizontally and determine its velocity after 0.7 seconds.

### Step 1: Identify the initial conditions
The body is projected horizontally with an initial velocity \( v_1 = 4\sqrt{2} \, \text{m/s} \). The initial vertical velocity \( v_{2i} = 0 \, \text{m/s} \) since it is projected horizontally.

### Step 2: Determine the vertical velocity after 0.7 seconds
The vertical component of the velocity can be calculated using the formula:
\[ v_{2} = v_{2i} + g \cdot t \]
Where:
- \( v_{2i} = 0 \, \text{m/s} \) (initial vertical velocity)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( t = 0.7 \, \text{s} \)

Substituting the values:
\[ v_{2} = 0 + 10 \cdot 0.7 = 7 \, \text{m/s} \]

### Step 3: Calculate the horizontal component of the velocity
The horizontal component of the velocity remains constant since there are no horizontal forces acting on the body. Thus:
\[ v_{1} = 4\sqrt{2} \, \text{m/s} \]

### Step 4: Calculate the resultant velocity
The resultant velocity \( v \) can be found using the Pythagorean theorem:
\[ v = \sqrt{v_{1}^2 + v_{2}^2} \]

Substituting the values:
\[ v = \sqrt{(4\sqrt{2})^2 + (7)^2} \]
Calculating \( (4\sqrt{2})^2 \):
\[ (4\sqrt{2})^2 = 16 \cdot 2 = 32 \]
Now substituting back into the equation:
\[ v = \sqrt{32 + 49} = \sqrt{81} = 9 \, \text{m/s} \]

### Conclusion
The velocity of the body after 0.7 seconds is \( 9 \, \text{m/s} \).

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