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 find the magnitude of the velocity of the body 0.2 seconds after it is projected, we will follow these steps:

### Step 1: Determine the horizontal and vertical displacement equations
The equations given for the horizontal and vertical displacements are:
- \( x = 2t \)
- \( y = 5t^2 \)

### Step 2: Calculate the horizontal velocity
The horizontal velocity \( V_x \) can be found by differentiating the horizontal displacement \( x \) with respect to time \( t \):
\[ V_x = \frac{dx}{dt} = \frac{d}{dt}(2t) = 2 \, \text{m/s} \]
This indicates that the horizontal velocity is constant at \( 2 \, \text{m/s} \).

### Step 3: Calculate the vertical velocity
The vertical velocity \( V_y \) can be found by differentiating the vertical displacement \( y \) with respect to time \( t \):
\[ V_y = \frac{dy}{dt} = \frac{d}{dt}(5t^2) = 10t \, \text{m/s} \]

### Step 4: Evaluate the vertical velocity at \( t = 0.2 \) seconds
Substituting \( t = 0.2 \) seconds into the equation for \( V_y \):
\[ V_y = 10(0.2) = 2 \, \text{m/s} \]

### Step 5: Calculate the magnitude of the total velocity
The magnitude of the total velocity \( V \) can be calculated using the Pythagorean theorem:
\[ V = \sqrt{V_x^2 + V_y^2} = \sqrt{(2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \, \text{m/s} \]

### 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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