A body is projected horizontally from a point above the ground. The motion of the body
is described by the equations x = 2 t and `y= 5 t^2` where x and y are the horizontal and vertical displacements (in m) respectively at time t. What is the magnitude of the velocity of the body 0.2 second after it is projected?

Options

A. `sqrt2 ms^(-1)`
B. `2 sqrt2 ms^(-1)`
C. ` 3.5 ms ^(-1)`
D. `5 ms ^(-1)`

Explanation

To solve the problem, we need to determine the magnitude of the velocity of the body 0.2 seconds after it is projected. The motion is described by the equations \( x = 2t \) and \( y = 5t^2 \).

### Step-by-Step Solution:

1. **Identify the equations of motion**:
- The horizontal displacement is given by \( x = 2t \).
- The vertical displacement is given by \( y = 5t^2 \).

2. **Find the horizontal component of velocity (\( v_x \))**:
- To find the horizontal velocity, differentiate \( x \) with respect to time \( t \):
\[ v_x = \frac{dx}{dt} = \frac{d(2t)}{dt} = 2 \, \text{m/s} \]
- The horizontal velocity \( v_x \) is constant and equal to \( 2 \, \text{m/s} \).

3. **Find the vertical component of velocity (\( v_y \))**:
- To find the vertical velocity, differentiate \( y \) with respect to time \( t \):
\[ v_y = \frac{dy}{dt} = \frac{d(5t^2)}{dt} = 10t \, \text{m/s} \]

4. **Calculate \( v_y \) at \( t = 0.2 \, \text{s} \)**:
- Substitute \( t = 0.2 \) into the equation for \( v_y \):
\[ v_y = 10(0.2) = 2 \, \text{m/s} \]

5. **Determine the magnitude of the total velocity (\( v \))**:
- The total velocity is found using the Pythagorean theorem since \( v_x \) and \( v_y \) are perpendicular:
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{(2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \, \text{m/s} \]

### Final Answer:
The magnitude of the velocity of the body 0.2 seconds after it is projected is \( 2\sqrt{2} \, \text{m/s} \).

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