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 parameters
The initial speed of the stone is given as \( V_0 = 10 \, \text{m/s} \) and the angle of projection is \( \theta_0 = 30^\circ \).

### Step 2: Calculate the initial velocity components
We need to resolve the initial velocity into its horizontal and vertical components:
- The horizontal component of the velocity \( V_{0x} \) is given by:
\[ V_{0x} = V_0 \cdot \cos(\theta_0) = 10 \cdot \cos(30^\circ) = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s} \]
- The vertical component of the velocity \( V_{0y} \) is given by:
\[ V_{0y} = V_0 \cdot \sin(\theta_0) = 10 \cdot \sin(30^\circ) = 10 \cdot \frac{1}{2} = 5 \, \text{m/s} \]

### Step 3: Calculate the displacement in the x-direction
Since there is no acceleration in the x-direction, the displacement \( x \) after time \( t \) can be calculated using the formula:
\[ x = V_{0x} \cdot t \]
Substituting the values:
\[ x = 5\sqrt{3} \cdot \frac{1}{2} = \frac{5\sqrt{3}}{2} \, \text{m} \]

### Step 4: Calculate the displacement in the y-direction
The displacement \( y \) in the y-direction can be calculated using the equation of motion:
\[ y = V_{0y} \cdot t - \frac{1}{2} g t^2 \]
where \( g \) (acceleration due to gravity) is approximately \( 10 \, \text{m/s}^2 \). Substituting the values:
\[ y = 5 \cdot \frac{1}{2} - \frac{1}{2} \cdot 10 \cdot \left(\frac{1}{2}\right)^2 \]
Calculating this step-by-step:
\[ y = \frac{5}{2} - \frac{1}{2} \cdot 10 \cdot \frac{1}{4} = \frac{5}{2} - \frac{10}{8} = \frac{5}{2} - \frac{5}{4} \]
Finding a common denominator:
\[ y = \frac{10}{4} - \frac{5}{4} = \frac{5}{4} \, \text{m} \]

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

### Step 6: Calculate the distance from the point of projection
To find the distance \( d \) from the point of projection to the position of the stone, we can use the Pythagorean theorem:
\[ d = \sqrt{x^2 + y^2} \]
Substituting the values:
\[ d = \sqrt{\left(\frac{5\sqrt{3}}{2}\right)^2 + \left(\frac{5}{4}\right)^2} \]
Calculating each term:
\[ = \sqrt{\frac{75}{4} + \frac{25}{16}} = \sqrt{\frac{300}{16} + \frac{25}{16}} = \sqrt{\frac{325}{16}} = \frac{5\sqrt{13}}{4} \, \text{m} \]

### Final Answer
Thus, the position of the stone with respect to the point of projection just after \( t = \frac{1}{2} \) seconds is:
\[ \left(\frac{5\sqrt{3}}{2}, \frac{5}{4}\right) \, \text{m} \]
And the distance from the point of projection to the stone is:
\[ \frac{5\sqrt{13}}{4} \, \text{m} \]