A body is projected at time t = 0 from a certain point on a planet’s surface with a certain velocity at a certain angle with the planet’s surface (assumed horizontal). The horizontal and vertical displacements x and y (in meters) respectively vary with time t (in second) as `x= 10 sqrt3t , y= 10t -t^2`
What is the magnitude and direction of the velocity with which the body is projected ?

Options

A. `20 ms^(-1)` at an angle of `30^@` with the horizontal
B. `20 ms^(-1)` at an angle `60^@` with the horizontal
C. `10 ms^(-1)` at an angle of `30^@` with the horizontal
D. `10 ms^(-1)` at an angle of `60^@` with horizontal

Explanation

To solve the problem, we need to find the magnitude and direction of the velocity with which the body is projected. The horizontal and vertical displacements are given by the equations:

- \( x = 10\sqrt{3}t \)
- \( y = 10t - t^2 \)

### Step 1: Find the velocity components
The velocity components in the horizontal and vertical directions can be found by taking the derivatives of the displacement equations with respect to time \( t \).

1. **Horizontal Velocity (\( V_x \))**:
\[ V_x = \frac{dx}{dt} = \frac{d}{dt}(10\sqrt{3}t) = 10\sqrt{3} \, \text{m/s} \]

2. **Vertical Velocity (\( V_y \))**:
\[ V_y = \frac{dy}{dt} = \frac{d}{dt}(10t - t^2) = 10 - 2t \, \text{m/s} \]

### Step 2: Calculate the velocity at \( t = 0 \)
To find the initial velocity with which the body is projected, we substitute \( t = 0 \) into the velocity equations.

1. **At \( t = 0 \)**:
- \( V_x = 10\sqrt{3} \, \text{m/s} \)
- \( V_y = 10 - 2(0) = 10 \, \text{m/s} \)

### Step 3: Calculate the magnitude of the resultant velocity
The magnitude of the resultant velocity \( V \) can be calculated using the Pythagorean theorem:
\[ V = \sqrt{V_x^2 + V_y^2} \]
Substituting the values:
\[ V = \sqrt{(10\sqrt{3})^2 + (10)^2} = \sqrt{300 + 100} = \sqrt{400} = 20 \, \text{m/s} \]

### Step 4: Calculate the direction of the velocity
The direction (angle \( \theta \)) of the velocity can be found using the tangent function:
\[ \tan(\theta) = \frac{V_y}{V_x} = \frac{10}{10\sqrt{3}} = \frac{1}{\sqrt{3}} \]
From this, we can find \( \theta \):
\[ \theta = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = 30^\circ \]

### Final Answer
The magnitude of the velocity with which the body is projected is \( 20 \, \text{m/s} \) at an angle of \( 30^\circ \) with the horizontal.

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