After one second the velocity of a projectile makes an angle of `45^(@)` with the horizontal. After another one more second its is travelling horizontally. The magnitude of its initial velocity and angle of projectile are
`(g=10m//s^(2))`

Options

A. `14.62m//s, 60^(@)`
B. `14.62m//s, tan^(-1)(2)`
C. `22.36m//s, tan^(-1)(2)`
D. `22.36 m//s, 60^(@)`

Answer

To solve the problem step by step, we will analyze the projectile motion based on the information provided.

### Step 1: Understanding the Problem
We know that after 1 second, the projectile makes an angle of 45° with the horizontal, and after 2 seconds, it is moving horizontally. We need to find the magnitude of the initial velocity and the angle of projection.

### Step 2: Define the Initial Velocity Components
Let the initial velocity be \( u \) and the angle of projection be \( \theta \).
- The horizontal component of the initial velocity is:
\[ u_x = u \cos \theta \]
- The vertical component of the initial velocity is:
\[ u_y = u \sin \theta \]

### Step 3: Analyze the Velocity After 1 Second
After 1 second, the velocity components are:
- Horizontal velocity (remains constant):
\[ v_x = u_x = u \cos \theta \]
- Vertical velocity after 1 second (considering gravity):
\[ v_y = u_y - g \cdot t = u \sin \theta - g \cdot 1 = u \sin \theta - 10 \]
Since the projectile makes a 45° angle with the horizontal:
\[ v_x = v_y \]
Thus, we have:
\[ u \cos \theta = u \sin \theta - 10 \quad \text{(1)} \]

### Step 4: Analyze the Velocity After 2 Seconds
After 2 seconds, the vertical component of the velocity becomes zero (the projectile is moving horizontally):
\[ v_y = u_y - g \cdot t = u \sin \theta - g \cdot 2 = u \sin \theta - 20 = 0 \]
From this, we can express:
\[ u \sin \theta = 20 \quad \text{(2)} \]

### Step 5: Substitute Equation (2) into Equation (1)
From equation (2):
\[ u \sin \theta = 20 \]
Substituting this into equation (1):
\[ u \cos \theta = 20 - 10 = 10 \quad \text{(3)} \]

### Step 6: Solve for \( u \) and \( \theta \)
Now we have two equations:
1. \( u \sin \theta = 20 \)
2. \( u \cos \theta = 10 \)

We can find \( u \) by squaring both equations and adding them:
\[ (u \sin \theta)^2 + (u \cos \theta)^2 = 20^2 + 10^2 \]
\[ u^2 (\sin^2 \theta + \cos^2 \theta) = 400 + 100 \]
Since \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ u^2 = 500 \quad \Rightarrow \quad u = \sqrt{500} = 10\sqrt{5} \approx 22.36 \, \text{m/s} \]

### Step 7: Find the Angle \( \theta \)
Now, we can find \( \theta \) using the ratios of the components:
\[ \tan \theta = \frac{u \sin \theta}{u \cos \theta} = \frac{20}{10} = 2 \]
Thus,
\[ \theta = \tan^{-1}(2) \]

### Final Results
The magnitude of the initial velocity is approximately \( 22.36 \, \text{m/s} \) and the angle of projection is \( \tan^{-1}(2) \).

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