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 follow the principles of projectile motion and trigonometry.

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

Using the definitions of sine and cosine:
- \( \tan \theta_0 = \frac{3}{4} \)
- This means that in a right triangle, the opposite side (perpendicular) is 3, and the adjacent side (base) is 4.
- The hypotenuse can be calculated using Pythagoras' theorem:
\[ \text{hypotenuse} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 \]

From this, we can find:
- \( \sin \theta_0 = \frac{3}{5} \)
- \( \cos \theta_0 = \frac{4}{5} \)

### Step 2: Calculate the horizontal and vertical components of the initial velocity
Using the initial velocity and the trigonometric functions:
- Horizontal component of velocity \( u_x = u \cos \theta_0 = 30 \cdot \frac{4}{5} = 24 \, \text{m/s} \)
- Vertical component of velocity \( u_y = u \sin \theta_0 = 30 \cdot \frac{3}{5} = 18 \, \text{m/s} \)

### Step 3: Determine the vertical velocity after 1 second
The vertical velocity changes due to gravity. The acceleration due to gravity \( g = 10 \, \text{m/s}^2 \) acts downwards. After 1 second, the vertical velocity \( v_y \) can be calculated as:
\[ v_y = u_y - g \cdot t = 18 - 10 \cdot 1 = 8 \, \text{m/s} \]

### Step 4: Determine the angle \( \theta \) after 1 second
After 1 second, the horizontal velocity remains constant:
- \( v_x = u_x = 24 \, \text{m/s} \)

Now, we can find the angle \( \theta \) that the resultant velocity makes with the horizontal:
\[ \tan \theta = \frac{v_y}{v_x} = \frac{8}{24} = \frac{1}{3} \]

### Step 5: Conclusion
Therefore, the value of \( \tan \theta \) after 1 second is \( \frac{1}{3} \).

### Final Answer
\[ \tan \theta = \frac{1}{3} \]

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