A piece of marble is projected from earth's surface with velocity of `19.6 sqrt(2) ms^(-1)` at `45^(@)`. 2 s later its velocity makes an angle `alpha` with horizontal, where `alpha` is

Options

A. `45^(@)`
B. `30^(@)`
C. `60^(@)`
D. `0^(@)`

Explanation

To solve the problem step by step, we will analyze the motion of the marble projected from the Earth's surface.

### Step 1: Identify the initial conditions
The marble is projected with an initial velocity \( u = 19.6 \sqrt{2} \, \text{m/s} \) at an angle of \( 45^\circ \) with respect to the horizontal.

### Step 2: Break down the initial velocity into components
The initial velocity can be resolved into horizontal and vertical components using trigonometric functions:
- Horizontal component \( u_x = u \cos(45^\circ) = 19.6 \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 19.6 \, \text{m/s} \)
- Vertical component \( u_y = u \sin(45^\circ) = 19.6 \sqrt{2} \cdot \frac{1}{\sqrt{2}} = 19.6 \, \text{m/s} \)

### Step 3: Calculate the vertical velocity after 2 seconds
The vertical velocity at any time \( t \) can be calculated using the equation:
\[ v_y = u_y - g t \]
where \( g \approx 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity. Substituting the values:
\[ v_y = 19.6 - 9.8 \cdot 2 = 19.6 - 19.6 = 0 \, \text{m/s} \]

### Step 4: Determine the horizontal velocity
The horizontal component of velocity remains constant throughout the motion since there is no horizontal acceleration (ignoring air resistance):
\[ v_x = u_x = 19.6 \, \text{m/s} \]

### Step 5: Calculate the resultant velocity after 2 seconds
After 2 seconds, the velocity components are:
- \( v_x = 19.6 \, \text{m/s} \)
- \( v_y = 0 \, \text{m/s} \)

The resultant velocity \( v \) can be calculated using the Pythagorean theorem:
\[ v = \sqrt{v_x^2 + v_y^2} = \sqrt{(19.6)^2 + (0)^2} = 19.6 \, \text{m/s} \]

### Step 6: Calculate the angle \( \alpha \) with the horizontal
The angle \( \alpha \) can be found using the tangent function:
\[ \tan(\alpha) = \frac{v_y}{v_x} = \frac{0}{19.6} = 0 \]
This implies that:
\[ \alpha = \tan^{-1}(0) = 0^\circ \]

### Final Answer
The angle \( \alpha \) that the velocity makes with the horizontal after 2 seconds is \( 0^\circ \).

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