A particle is projected with a speed u. If after 2 seconds of projection it is found to be making an angle of `45^@` with the horizontal and `0^@` after 3 sec, then:

Options

A. Angle of projection is `tan^(-1) 3`
B. Angle of projection is `tan^(-1) (1//3)`
C. Speed of projection `= 30 sqrt2m//s`
D. Speed of projection is 30 m/s

Explanation

Let's solve the problem step by step.

### Given:
- A particle is projected with an initial speed \( u \).
- After \( t = 2 \) seconds, the particle makes an angle of \( 45^\circ \) with the horizontal.
- After \( t = 3 \) seconds, the particle is at its maximum height, making an angle of \( 0^\circ \) with the horizontal.

### Step 1: Analyze the motion at \( t = 3 \) seconds
At \( t = 3 \) seconds, the particle is at its maximum height. This means the vertical component of its velocity is zero.

Using the vertical motion equation:
\[ v_y = u \sin(\theta) - g t \]
Setting \( v_y = 0 \) at \( t = 3 \):
\[ 0 = u \sin(\theta) - 3g \]
Thus,
\[ u \sin(\theta) = 3g \quad \text{(1)} \]

### Step 2: Analyze the motion at \( t = 2 \) seconds
At \( t = 2 \) seconds, the angle with the horizontal is \( 45^\circ \). Therefore, the vertical and horizontal components of the velocity are equal.

Using the vertical motion equation:
\[ v_y = u \sin(\theta) - g t \]
At \( t = 2 \):
\[ v_y = u \sin(\theta) - 2g \]
Since the angle is \( 45^\circ \), we have:
\[ v_y = v \sin(45^\circ) = \frac{v}{\sqrt{2}} \quad \text{(2)} \]
And for the horizontal component:
\[ v_x = u \cos(\theta) \quad \text{(3)} \]
Since \( v_x = v \cos(45^\circ) = \frac{v}{\sqrt{2}} \), we can equate the two:
\[ u \cos(\theta) = \frac{v}{\sqrt{2}} \quad \text{(4)} \]

### Step 3: Relate equations (1) and (4)
From equation (1):
\[ u \sin(\theta) = 3g \]
From equation (4):
\[ u \cos(\theta) = \frac{v}{\sqrt{2}} \]
Now we need to find \( v \). At \( t = 3 \):
\[ v = \sqrt{(u \cos(\theta))^2 + (0)^2} = u \cos(\theta) \]
Substituting \( v \) into equation (4):
\[ u \cos(\theta) = \frac{u \cos(\theta)}{\sqrt{2}} \]
This implies:
\[ v = u \cos(\theta) = \sqrt{2} \cdot u \cos(\theta) \]

### Step 4: Solve for \( u \) and \( \theta \)
Now we can find \( \tan(\theta) \):
\[ \tan(\theta) = \frac{u \sin(\theta)}{u \cos(\theta)} = \frac{3g}{\frac{v}{\sqrt{2}}} \]
Using \( g = 10 \, \text{m/s}^2 \):
\[ \tan(\theta) = \frac{3 \cdot 10}{\frac{v}{\sqrt{2}}} \]
From the previous equations, we can find \( u \) and \( \theta \) using the Pythagorean identity:
\[ u^2 = (u \sin(\theta))^2 + (u \cos(\theta))^2 \]
Substituting the values from equations (1) and (4) into this identity will yield the values of \( u \) and \( \theta \).

### Final Calculation
1. From equation (1), \( u \sin(\theta) = 30 \).
2. From equation (4), \( u \cos(\theta) = 10 \).
3. Using \( \sin^2(\theta) + \cos^2(\theta) = 1 \):
\[ \left(\frac{30}{u}\right)^2 + \left(\frac{10}{u}\right)^2 = 1 \]
\[ \frac{900 + 100}{u^2} = 1 \implies u^2 = 1000 \implies u = 10\sqrt{10} \]

### Conclusion
The angle of projection \( \theta \) can be found using:
\[ \tan(\theta) = \frac{3}{1} \implies \theta = \tan^{-1}(3) \]