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 find the angular momentum of the particle about the point of projection when it is at the highest point, we can follow these steps:

### Step 1: Identify the parameters
- 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 vertical and horizontal components of the initial velocity
- The horizontal component of velocity (u_x) = u * cos(θ)
\[ u_x = 20 \cdot \cos(60°) = 20 \cdot \frac{1}{2} = 10 \, \text{m/s} \]

- The vertical component of velocity (u_y) = u * sin(θ)
\[ u_y = 20 \cdot \sin(60°) = 20 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3} \, \text{m/s} \]

### Step 3: Calculate the time taken to reach the highest point
At the highest point, the vertical component of the velocity becomes zero. We can use the formula:
\[ v_y = u_y - g t \]
Setting \(v_y = 0\):
\[ 0 = 10\sqrt{3} - 10t \implies t = \sqrt{3} \, \text{s} \]

### Step 4: Calculate the horizontal distance (range) traveled to the highest point
The horizontal distance (R) traveled in time t is given by:
\[ R = u_x \cdot t = 10 \cdot \sqrt{3} \, \text{m} \]

### Step 5: Calculate the height (h) at the highest point
The height can be calculated using:
\[ h = u_y t - \frac{1}{2} g t^2 \]
Substituting the values:
\[ h = 10\sqrt{3} \cdot \sqrt{3} - \frac{1}{2} \cdot 10 \cdot (\sqrt{3})^2 = 30 - 15 = 15 \, \text{m} \]

### Step 6: Calculate the angular momentum (L) at the highest point
The angular momentum about the point of projection is given by:
\[ L = m \cdot v \cdot r_{\perpendicular} \]
Where \(v\) is the horizontal component of velocity at the highest point (which remains the same as \(u_x\)), and \(r_{\perpendicular}\) is the distance from the point of projection to the line of action of the velocity vector.

Here, \(r_{\perpendicular} = h\) (the height) at the highest point:
\[ L = m \cdot u_x \cdot h = 1 \cdot 10 \cdot 15 = 150 \, \text{kg m}^2/\text{s} \]

### Final Answer
The angular momentum of the particle about the point of projection when it is at the highest point is \(150 \, \text{kg m}^2/\text{s}\).

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