A body is projected with velocity `u` at an angle of projection `theta` with the horizontal. The direction of velocity of the body makes angle `30^@` with the horizontal at `t = 2 s` and then after `1 s` it reaches the maximum height. Then

Options

A. a. `u = 20 sqrt(3) m s^-1`,
B. b. `theta = 60^@`
C. c. `theta = 30^@`
D. d. `u = 10 sqrt(3) m s^-1`

Explanation

To solve the problem step by step, we will analyze the projectile motion of the body and derive the required values.

### Step 1: Understand the given information
We know that:
- The body is projected with an initial velocity \( u \) at an angle \( \theta \) with the horizontal.
- At \( t = 2 \) seconds, the direction of the velocity makes an angle of \( 30^\circ \) with the horizontal.
- After \( 1 \) second from that moment (i.e., at \( t = 3 \) seconds), the body reaches its maximum height.

### Step 2: Determine the total time of ascent
Since the body reaches its maximum height at \( t = 3 \) seconds, the total time of ascent is:
\[ T = 3 \text{ seconds} \]

### Step 3: Relate time of ascent to initial vertical velocity
The time of ascent \( T \) is given by the formula:
\[ T = \frac{u \sin \theta}{g} \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)). Thus, we can write:
\[ 3 = \frac{u \sin \theta}{10} \]
From this, we can express \( u \sin \theta \):
\[ u \sin \theta = 30 \quad \text{(Equation 1)} \]

### Step 4: Analyze the velocity components at \( t = 2 \) seconds
At \( t = 2 \) seconds, the horizontal and vertical components of the velocity can be expressed as:
- Horizontal component \( V_x = u \cos \theta \)
- Vertical component \( V_y = u \sin \theta - g t \)

Substituting \( t = 2 \) seconds into the vertical component:
\[ V_y = u \sin \theta - 20 \quad \text{(since } g \approx 10 \text{ m/s}^2\text{)} \]

### Step 5: Use the angle of velocity at \( t = 2 \) seconds
Since the angle of the velocity makes \( 30^\circ \) with the horizontal, we can use the tangent function:
\[ \tan(30^\circ) = \frac{V_y}{V_x} \]
Given \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \), we can write:
\[ \frac{V_y}{V_x} = \frac{1}{\sqrt{3}} \]
Substituting the expressions for \( V_y \) and \( V_x \):
\[ \frac{u \sin \theta - 20}{u \cos \theta} = \frac{1}{\sqrt{3}} \]

### Step 6: Cross-multiply and simplify
Cross-multiplying gives:
\[ \sqrt{3}(u \sin \theta - 20) = u \cos \theta \]
Expanding this:
\[ \sqrt{3} u \sin \theta - 20\sqrt{3} = u \cos \theta \]
Rearranging gives us:
\[ \sqrt{3} u \sin \theta - u \cos \theta = 20\sqrt{3} \]
Factoring out \( u \):
\[ u (\sqrt{3} \sin \theta - \cos \theta) = 20\sqrt{3} \quad \text{(Equation 2)} \]

### Step 7: Solve the equations
Now we have two equations:
1. \( u \sin \theta = 30 \)
2. \( u (\sqrt{3} \sin \theta - \cos \theta) = 20\sqrt{3} \)

Substituting \( u \sin \theta = 30 \) into Equation 2:
\[ u \left(\sqrt{3} \cdot \frac{30}{u} - \cos \theta\right) = 20\sqrt{3} \]
This simplifies to:
\[ 30\sqrt{3} - u \cos \theta = 20\sqrt{3} \]
Thus:
\[ u \cos \theta = 10\sqrt{3} \]

### Step 8: Find \( u \) and \( \theta \)
Now we have:
1. \( u \sin \theta = 30 \)
2. \( u \cos \theta = 10\sqrt{3} \)

Dividing these two equations:
\[ \frac{u \sin \theta}{u \cos \theta} = \frac{30}{10\sqrt{3}} \implies \tan \theta = \frac{3}{\sqrt{3}} = \sqrt{3} \]
Thus:
\[ \theta = 60^\circ \]

### Step 9: Calculate \( u \)
Using \( u \sin 60^\circ = 30 \):
\[ u \cdot \frac{\sqrt{3}}{2} = 30 \implies u = \frac{30 \cdot 2}{\sqrt{3}} = 20\sqrt{3} \]

### Final Answer
Thus, we find:
- \( u = 20\sqrt{3} \)
- \( \theta = 60^\circ \)