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

To solve the problem step by step, we will analyze the motion of the particle projected at an angle and use the information given about its velocity at specific times.

### Step 1: Understand the components of motion
The particle is projected with an initial speed \( u \) at an angle \( \theta \). The initial velocity can be broken down into horizontal and vertical components:
- Horizontal component: \( u_x = u \cos \theta \)
- Vertical component: \( u_y = u \sin \theta \)

### Step 2: Analyze the vertical motion at \( t = 3 \) seconds
At \( t = 3 \) seconds, the particle is at an angle of \( 0^\circ \) with the horizontal, which means its vertical component of velocity \( v_y \) is \( 0 \):
\[ v_y = u_y - g t = 0 \]
Substituting \( t = 3 \) seconds:
\[ u \sin \theta - g \cdot 3 = 0 \]
Rearranging gives:
\[ u \sin \theta = 3g \]
Using \( g \approx 10 \, \text{m/s}^2 \):
\[ u \sin \theta = 30 \, \text{m/s} \quad \text{(1)} \]

### Step 3: Analyze the vertical motion at \( t = 2 \) seconds
At \( t = 2 \) seconds, the angle with the horizontal is \( 45^\circ \). The tangent of the angle is given by:
\[ \tan(45^\circ) = 1 = \frac{v_y}{u_x} \]
The vertical component of velocity at \( t = 2 \) seconds is:
\[ v_y = u \sin \theta - g \cdot 2 = u \sin \theta - 20 \]
Setting this equal to \( u \cos \theta \) (since \( \tan(45^\circ) = 1 \)):
\[ u \sin \theta - 20 = u \cos \theta \]
Rearranging gives:
\[ u \sin \theta - u \cos \theta = 20 \]
Factoring out \( u \):
\[ u (\sin \theta - \cos \theta) = 20 \quad \text{(2)} \]

### Step 4: Solve the equations
Now we have two equations:
1. \( u \sin \theta = 30 \)
2. \( u (\sin \theta - \cos \theta) = 20 \)

From equation (1), we can express \( \sin \theta \):
\[ \sin \theta = \frac{30}{u} \]
Substituting this into equation (2):
\[ u \left(\frac{30}{u} - \cos \theta\right) = 20 \]
This simplifies to:
\[ 30 - u \cos \theta = 20 \]
Thus:
\[ u \cos \theta = 10 \quad \text{(3)} \]

### Step 5: Find \( \tan \theta \)
Now we have:
- From (1): \( u \sin \theta = 30 \)
- From (3): \( u \cos \theta = 10 \)

Now we can find \( \tan \theta \):
\[ \tan \theta = \frac{u \sin \theta}{u \cos \theta} = \frac{30}{10} = 3 \]
Thus:
\[ \theta = \tan^{-1}(3) \]

### Step 6: Find the initial speed \( u \)
To find \( u \), we can use the Pythagorean theorem:
\[ u^2 = (u \sin \theta)^2 + (u \cos \theta)^2 \]
Substituting the values:
\[ u^2 = 30^2 + 10^2 = 900 + 100 = 1000 \]
Thus:
\[ u = \sqrt{1000} = 10\sqrt{10} \, \text{m/s} \]

### Final Answers
- The angle of projection \( \theta = \tan^{-1}(3) \)
- The speed of projection \( u = 10\sqrt{10} \, \text{m/s} \)