A body is projected with an initial elocity 20m/s at 60° to the horizontal. The displacement after 2 s is

Options

A. `20[i+(sqrt(3)-1)i]`
B. `20[1-(sqrt(3)-1)j]`
C. `10[i+(sqrt(3)+1)j]`
D. `10[1+(sqrt(3)-1)]`

Answer

To solve the problem step by step, we need to find the displacement of a body projected with an initial velocity of 20 m/s at an angle of 60° to the horizontal after 2 seconds.

### Step 1: Determine the initial velocity components
The initial velocity \( U \) is given as 20 m/s. We need to find the horizontal and vertical components of this velocity.

- **Horizontal component \( U_H \)**:
\[ U_H = U \cos(60°) = 20 \cos(60°) = 20 \times \frac{1}{2} = 10 \, \text{m/s} \]

- **Vertical component \( U_V \)**:
\[ U_V = U \sin(60°) = 20 \sin(60°) = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{m/s} \]

### Step 2: Calculate horizontal displacement after 2 seconds
Since there is no acceleration in the horizontal direction, the horizontal displacement \( S_H \) can be calculated using the formula:
\[ S_H = U_H \times T \]
Substituting the values:
\[ S_H = 10 \, \text{m/s} \times 2 \, \text{s} = 20 \, \text{m} \]

### Step 3: Calculate vertical displacement after 2 seconds
For vertical displacement \( S_V \), we use the equation:
\[ S_V = U_V \times T - \frac{1}{2} g T^2 \]
Where \( g \) (acceleration due to gravity) is approximately \( 10 \, \text{m/s}^2 \). Substituting the values:
\[ S_V = (10\sqrt{3} \, \text{m/s}) \times 2 \, \text{s} - \frac{1}{2} \times 10 \, \text{m/s}^2 \times (2 \, \text{s})^2 \]
Calculating each term:
\[ S_V = 20\sqrt{3} - \frac{1}{2} \times 10 \times 4 \]
\[ S_V = 20\sqrt{3} - 20 \]

### Step 4: Combine horizontal and vertical displacements
Now we can express the total displacement vector \( \vec{S} \) in terms of its components:
\[ \vec{S} = S_H \hat{i} + S_V \hat{j} \]
Substituting the values we found:
\[ \vec{S} = 20 \hat{i} + (20\sqrt{3} - 20) \hat{j} \]

### Step 5: Final expression for displacement
Thus, the displacement after 2 seconds can be written as:
\[ \vec{S} = 20 \hat{i} + (20\sqrt{3} - 20) \hat{j} \]