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 find the magnitude and direction of the velocity with which the body is projected, we will follow these steps:

### Step 1: Differentiate the displacement equations
We are given the horizontal and vertical displacements as:
- \( x(t) = 10\sqrt{3}t \)
- \( y(t) = 10t - t^2 \)

To find the velocity components, we differentiate these equations with respect to time \( t \).

### Step 2: Calculate the horizontal velocity component
Differentiating \( x(t) \):
\[ v_x = \frac{dx}{dt} = \frac{d}{dt}(10\sqrt{3}t) = 10\sqrt{3} \, \text{m/s} \]

### Step 3: Calculate the vertical velocity component
Differentiating \( y(t) \):
\[ v_y = \frac{dy}{dt} = \frac{d}{dt}(10t - t^2) = 10 - 2t \, \text{m/s} \]

### Step 4: Evaluate the vertical velocity at \( t = 0 \)
At \( t = 0 \):
\[ v_y(0) = 10 - 2(0) = 10 \, \text{m/s} \]

### Step 5: Find the magnitude of the velocity
The magnitude of the 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 6: Find the direction of the velocity
The direction (angle \( \theta \)) can be found using the tangent function:
\[ \tan(\theta) = \frac{v_y}{v_x} \]
Substituting the values:
\[ \tan(\theta) = \frac{10}{10\sqrt{3}} = \frac{1}{\sqrt{3}} \]
Thus,
\[ \theta = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = 30^\circ \]

### Conclusion
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.