A boy throws a stone with a speed of `V_(0) = 10` m/sec at an angle of `theta_(0)= 30^(@)` to the horizontal Find the position of the stone wrt the point of projection just after a time t = 1/2 sec.

Answer

To solve the problem of finding the position of the stone with respect to the point of projection just after a time \( t = \frac{1}{2} \) seconds, we can break it down into the following steps:

### Step 1: Identify the initial conditions
- Initial speed \( V_0 = 10 \, \text{m/s} \)
- Angle of projection \( \theta_0 = 30^\circ \)

### Step 2: Calculate the initial velocity components
Using trigonometric functions, we can find the horizontal and vertical components of the initial velocity:
- Horizontal component \( V_{0x} = V_0 \cos(\theta_0) = 10 \cos(30^\circ) \)
- Vertical component \( V_{0y} = V_0 \sin(\theta_0) = 10 \sin(30^\circ) \)

Calculating these:
- \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \)
- \( \sin(30^\circ) = \frac{1}{2} \)

Thus:
- \( V_{0x} = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s} \)
- \( V_{0y} = 10 \times \frac{1}{2} = 5 \, \text{m/s} \)

### Step 3: Calculate the position after \( t = \frac{1}{2} \) seconds
The position can be calculated using the equations of motion. The horizontal and vertical displacements can be calculated separately.

#### Horizontal displacement \( x \):
\[ x = V_{0x} \cdot t \]
Substituting the values:
\[ x = 5\sqrt{3} \cdot \frac{1}{2} = \frac{5\sqrt{3}}{2} \, \text{m} \]

#### Vertical displacement \( y \):
The vertical displacement is affected by gravity. The equation for vertical displacement is:
\[ y = V_{0y} \cdot t - \frac{1}{2} g t^2 \]
Where \( g \approx 9.81 \, \text{m/s}^2 \). Substituting the values:
\[ y = 5 \cdot \frac{1}{2} - \frac{1}{2} \cdot 9.81 \cdot \left(\frac{1}{2}\right)^2 \]
Calculating:
\[ y = 2.5 - \frac{1}{2} \cdot 9.81 \cdot \frac{1}{4} = 2.5 - \frac{9.81}{8} \approx 2.5 - 1.22625 \approx 1.27375 \, \text{m} \]

### Step 4: Final position of the stone
The position of the stone with respect to the point of projection after \( t = \frac{1}{2} \) seconds is:
\[ \text{Position} = \left( \frac{5\sqrt{3}}{2}, 1.27375 \right) \, \text{m} \]

### Summary of the solution:
- Horizontal displacement \( x \approx 4.33 \, \text{m} \) (since \( \sqrt{3} \approx 1.732 \))
- Vertical displacement \( y \approx 1.27 \, \text{m} \)