A particle is projected with a velocity of 30 m/s at an angle `theta` with the horizontal . Where ` theta = tan^(-1) (3/4)` . After 1 second, direction of motion of the particle makes an angle ` alpha` with the horizontal then ` alpha` is given by

Options

A. `tan alpha = 1/3`
B. `tan alpha = 1/2`
C. `tan alpha = 2`
D. ` tan alpha= 1`

Answer

To solve the problem, we need to find the angle \( \alpha \) that the particle makes with the horizontal after 1 second of its motion. Here’s a step-by-step breakdown of the solution:

### Step 1: Determine the angle \( \theta \)
Given that:
\[ \theta = \tan^{-1}\left(\frac{3}{4}\right) \]
We can visualize this as a right triangle where the opposite side is 3 and the adjacent side is 4. The hypotenuse can be calculated using the Pythagorean theorem:
\[ \text{Hypotenuse} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]
Thus, we can find:
\[ \sin \theta = \frac{3}{5}, \quad \cos \theta = \frac{4}{5} \]

### Step 2: Calculate the initial velocity components
The initial velocity \( u \) is given as 30 m/s. The horizontal and vertical components of the initial velocity are:
\[ u_x = u \cos \theta = 30 \cdot \frac{4}{5} = 24 \, \text{m/s} \]
\[ u_y = u \sin \theta = 30 \cdot \frac{3}{5} = 18 \, \text{m/s} \]

### Step 3: Analyze the motion after 1 second
After 1 second, the horizontal component of the velocity remains constant (since there is no horizontal acceleration):
\[ v_x = u_x = 24 \, \text{m/s} \]

In the vertical direction, the velocity changes due to gravity:
\[ v_y = u_y - g \cdot t = 18 - 10 \cdot 1 = 8 \, \text{m/s} \]

### Step 4: Find the angle \( \alpha \)
The angle \( \alpha \) can be found using the tangent function:
\[ \tan \alpha = \frac{v_y}{v_x} = \frac{8}{24} = \frac{1}{3} \]

### Step 5: Calculate \( \alpha \)
To find \( \alpha \), we take the arctangent:
\[ \alpha = \tan^{-1}\left(\frac{1}{3}\right) \]

### Final Answer
Thus, the angle \( \alpha \) is given by:
\[ \alpha = \tan^{-1}\left(\frac{1}{3}\right) \]

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