A ball is projected from the ground at angle `theta` with the horizontal. After `1 s`, it is moving at angle `45^@` with the horizontal and after `2 s` it is moving horizontally. What is the velocity of projection of the ball ?

Options

A. `10 sqrt(3) ms^-1`
B. `20 sqrt(3) ms^-1`
C. `10 sqrt(5) ms^-1`
D. `20 sqrt(2) ms^-1`

Explanation

To solve the problem, we need to analyze the motion of the ball projected at an angle θ with the horizontal. We will use the information provided about the ball's motion at specific times to determine the velocity of projection.

### Step-by-Step Solution:

1. **Define the Initial Velocity Components**:
Let the initial velocity of projection be \( \mathbf{u} = u_x \hat{i} + u_y \hat{j} \), where \( u_x \) is the horizontal component and \( u_y \) is the vertical component of the initial velocity.

2. **Analyze the Motion After 1 Second**:
After 1 second, the ball is moving at an angle of 45 degrees with the horizontal. The tangent of the angle is given by:
\[ \tan(45^\circ) = 1 = \frac{v_y}{v_x} \]
where \( v_y \) and \( v_x \) are the vertical and horizontal components of the velocity after 1 second, respectively.

The vertical component of velocity after 1 second can be expressed as:
\[ v_y = u_y - g \cdot t = u_y - 10 \cdot 1 = u_y - 10 \]
The horizontal component remains constant:
\[ v_x = u_x \]
Setting the two equal gives:
\[ u_y - 10 = u_x \quad \text{(1)} \]

3. **Analyze the Motion After 2 Seconds**:
After 2 seconds, the ball is moving horizontally, which means the vertical component of the velocity is zero:
\[ v_y = u_y - g \cdot t = u_y - 10 \cdot 2 = u_y - 20 = 0 \]
From this, we can solve for \( u_y \):
\[ u_y = 20 \quad \text{(2)} \]

4. **Substitute \( u_y \) into Equation (1)**:
Now, substituting \( u_y = 20 \) into equation (1):
\[ 20 - 10 = u_x \implies u_x = 10 \]

5. **Calculate the Magnitude of the Initial Velocity**:
The initial velocity \( \mathbf{u} \) can now be expressed as:
\[ \mathbf{u} = 10 \hat{i} + 20 \hat{j} \]
The magnitude of the initial velocity \( u \) is given by:
\[ u = \sqrt{u_x^2 + u_y^2} = \sqrt{10^2 + 20^2} = \sqrt{100 + 400} = \sqrt{500} = 10\sqrt{5} \]

### Final Answer:
The velocity of projection of the ball is \( 10\sqrt{5} \, \text{m/s} \).