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 \(\theta\) with the horizontal. We will use the information given about the angles of motion at specific times to derive the initial velocity of projection.

### Step-by-step Solution:

1. **Understanding the Motion**:
The ball is projected from the ground at an angle \(\theta\). After \(1\) second, it makes an angle of \(45^\circ\) with the horizontal, and after \(2\) seconds, it moves horizontally.

2. **Using the Angle at \(t = 1\) second**:
At \(t = 1\) second, the angle of the velocity vector is \(45^\circ\). This means that the vertical and horizontal components of the velocity are equal at this point.
\[ \tan(45^\circ) = 1 = \frac{V_y}{V_x} \]
Here, \(V_y\) is the vertical component of the velocity at \(t = 1\) second, and \(V_x\) is the horizontal component.

The vertical component of the velocity can be expressed as:
\[ V_y = U \sin \theta - g \cdot t \]
Substituting \(t = 1\):
\[ V_y = U \sin \theta - g \]

The horizontal component remains constant:
\[ V_x = U \cos \theta \]

Setting \(V_y = V_x\) gives:
\[ U \sin \theta - g = U \cos \theta \]
Rearranging this, we get:
\[ U \sin \theta - U \cos \theta = g \]
\[ U (\sin \theta - \cos \theta) = g \quad \text{(Equation 1)} \]

3. **Using the Angle at \(t = 2\) seconds**:
At \(t = 2\) seconds, the ball is moving horizontally, which means that the vertical component of the velocity is zero:
\[ V_y = U \sin \theta - g \cdot 2 = 0 \]
Rearranging gives:
\[ U \sin \theta = 2g \quad \text{(Equation 2)} \]

4. **Solving the Equations**:
Now we have two equations:
- From Equation 1: \(U (\sin \theta - \cos \theta) = g\)
- From Equation 2: \(U \sin \theta = 2g\)

From Equation 2, we can express \(U\):
\[ U = \frac{2g}{\sin \theta} \]

Substituting \(U\) into Equation 1:
\[ \frac{2g}{\sin \theta} (\sin \theta - \cos \theta) = g \]
Dividing both sides by \(g\) (assuming \(g \neq 0\)):
\[ \frac{2(\sin \theta - \cos \theta)}{\sin \theta} = 1 \]
Rearranging gives:
\[ 2(\sin \theta - \cos \theta) = \sin \theta \]
\[ 2\sin \theta - 2\cos \theta = \sin \theta \]
\[ \sin \theta = 2\cos \theta \]
Dividing both sides by \(\cos \theta\) gives:
\[ \tan \theta = 2 \]

5. **Finding the Initial Velocity**:
Now substituting \(\tan \theta = 2\) back into Equation 2:
\[ U \sin \theta = 2g \]
We can express \(\sin \theta\) in terms of \(\tan \theta\):
\[ \sin \theta = \frac{2}{\sqrt{5}} \quad \text{and} \quad \cos \theta = \frac{1}{\sqrt{5}} \]
Thus,
\[ U = \frac{2g}{\frac{2}{\sqrt{5}}} = g\sqrt{5} \]

### Final Answer:
The velocity of projection of the ball is:
\[ U = g\sqrt{5} \text{ m/s} \]