A particle is projected with a velocity of `30 ms^(-1)`, at an angle of `theta_(0) = tan^(-1) ((3)/(4))`. After 1s, the particle is moving at an angle `theta` to the horizontal, where `tan theta` will be equal to (Take, `g = 10 ms^(-2))`

Options

A. 1
B. 2
C. `(1)/(2)`
D. `(1)/(3)`

Explanation

To solve the problem step by step, we will analyze the motion of the particle projected at an angle and find the required angle after 1 second.

### Step 1: Determine the initial components of velocity
The initial velocity \( u = 30 \, \text{m/s} \) is given, and the angle of projection \( \theta_0 = \tan^{-1}\left(\frac{3}{4}\right) \).

We need to find the horizontal and vertical components of the initial velocity:
- The horizontal component \( u_x = u \cos \theta_0 \)
- The vertical component \( u_y = u \sin \theta_0 \)

Using the trigonometric values:
- \( \cos \theta_0 = \frac{4}{5} \)
- \( \sin \theta_0 = \frac{3}{5} \)

Calculating:
\[ u_x = 30 \cdot \frac{4}{5} = 24 \, \text{m/s} \]
\[ u_y = 30 \cdot \frac{3}{5} = 18 \, \text{m/s} \]

### Step 2: Calculate the vertical velocity after 1 second
The vertical velocity \( v_y \) after time \( t = 1 \, \text{s} \) can be calculated using the equation:
\[ v_y = u_y - g t \]
where \( g = 10 \, \text{m/s}^2 \).

Substituting the values:
\[ v_y = 18 - 10 \cdot 1 = 18 - 10 = 8 \, \text{m/s} \]

### Step 3: Determine the horizontal velocity
The horizontal velocity \( v_x \) remains constant since there is no horizontal acceleration:
\[ v_x = u_x = 24 \, \text{m/s} \]

### Step 4: Calculate the angle \( \theta \)
Now, we need to find \( \tan \theta \):
\[ \tan \theta = \frac{v_y}{v_x} = \frac{8}{24} = \frac{1}{3} \]

### Conclusion
Thus, the value of \( \tan \theta \) is \( \frac{1}{3} \).