In the above problem the displacement after 2 s is:

Options

A. `30sqrt(3)hati + 30hatj`
B. `60sqrt(3)hati + 40 hatj`
C. `10sqrt(3)hati + 10 hatj`
D. `40sqrt(3)hati + 40 hatj`

Answer

To find the displacement of the ball after 2 seconds when thrown at a speed of 20 m/s at an angle of 30 degrees with the horizontal, we can follow these steps:

### Step 1: Resolve the initial velocity into horizontal and vertical components.
The initial velocity \( u \) is given as 20 m/s and the angle \( \theta \) is 30 degrees.

- The horizontal component \( u_x \) is given by:
\[ u_x = u \cdot \cos(\theta) = 20 \cdot \cos(30^\circ) \]
- The vertical component \( u_y \) is given by:
\[ u_y = u \cdot \sin(\theta) = 20 \cdot \sin(30^\circ) \]

### Step 2: Calculate the horizontal and vertical components.
Using the values of \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \) and \( \sin(30^\circ) = \frac{1}{2} \):

- Calculate \( u_x \):
\[ u_x = 20 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{m/s} \]
- Calculate \( u_y \):
\[ u_y = 20 \cdot \frac{1}{2} = 10 \, \text{m/s} \]

### Step 3: Calculate the horizontal displacement after 2 seconds.
The horizontal displacement \( x \) after time \( t \) is given by:
\[ x = u_x \cdot t \]
Substituting \( t = 2 \, \text{s} \):
\[ x = 10\sqrt{3} \cdot 2 = 20\sqrt{3} \, \text{m} \]

### Step 4: Calculate the vertical displacement after 2 seconds.
The vertical displacement \( y \) after time \( t \) is given by:
\[ y = u_y \cdot t - \frac{1}{2} g t^2 \]
Where \( g = 10 \, \text{m/s}^2 \). Substituting the values:
\[ y = 10 \cdot 2 - \frac{1}{2} \cdot 10 \cdot (2)^2 \]
\[ y = 20 - \frac{1}{2} \cdot 10 \cdot 4 \]
\[ y = 20 - 20 = 0 \, \text{m} \]

### Step 5: Calculate the resultant displacement.
The resultant displacement \( d \) can be calculated using the Pythagorean theorem:
\[ d = \sqrt{x^2 + y^2} \]
Substituting the values:
\[ d = \sqrt{(20\sqrt{3})^2 + 0^2} = \sqrt{1200} = 20\sqrt{3} \, \text{m} \]

### Final Answer:
The displacement after 2 seconds is \( 20\sqrt{3} \, \text{m} \).

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