A particle of mass 1kg has been thrown with initial speed 20 `m/s` making an angle 60° with the horizontal ground. the angular momentum of the particle about point of projection when the projectile is at highest point is (g=10m/s^2)

Options

A. `150 kg m^2/s`
B. `300 kg m^2/s`
C. Zero
D. `100 kg m^2/s`

Answer

To solve the problem of finding the angular momentum of a particle at its highest point in projectile motion, we can follow these steps:

### Step 1: Identify the given values
- Mass of the particle (m) = 1 kg
- Initial speed (u) = 20 m/s
- Angle of projection (θ) = 60°
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Calculate the horizontal and vertical components of the initial velocity
- The horizontal component of the initial velocity (u_x) is given by:
\[ u_x = u \cdot \cos(θ) = 20 \cdot \cos(60°) = 20 \cdot \frac{1}{2} = 10 \, \text{m/s} \]
- The vertical component of the initial velocity (u_y) is given by:
\[ u_y = u \cdot \sin(θ) = 20 \cdot \sin(60°) = 20 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{m/s} \]

### Step 3: Determine the maximum height (H_max)
The maximum height for a projectile is given by the formula:
\[ H_{max} = \frac{u_y^2}{2g} \]
Substituting the values:
\[ H_{max} = \frac{(10\sqrt{3})^2}{2 \cdot 10} = \frac{100 \cdot 3}{20} = 15 \, \text{m} \]

### Step 4: Calculate the angular momentum (L) at the highest point
At the highest point, the vertical component of the velocity is zero, and only the horizontal component remains. The angular momentum about the point of projection is given by:
\[ L = m \cdot v \cdot r \]
Where:
- \( v = u_x = 10 \, \text{m/s} \) (horizontal component)
- \( r = H_{max} = 15 \, \text{m} \) (perpendicular distance from the line of action of velocity to the point of projection)

Now substituting the values:
\[ L = 1 \cdot 10 \cdot 15 = 150 \, \text{kg m}^2/\text{s} \]

### Final Answer
The angular momentum of the particle about the point of projection when the projectile is at its highest point is:
\[ \boxed{150 \, \text{kg m}^2/\text{s}} \]