A ball of mass 160 g is thrown up at an angle of `60^(@)` to the horizontal at a speed of `10ms^(-1)` . The angular momentum of the ball at highest point of the trajectory with respect to the point from which the ball is thrown is nearly `(g=10ms^(2))`

Options

A. `1.73 kg m^(2)//s`
B. `3.0 kg m^(2) //s`
C. `3.46 kg m^(2) //s`
D. `6.0 kg m^(2)//s`

Explanation

To find the angular momentum of the ball at the highest point of its trajectory with respect to the point from which it was thrown, we can follow these steps:

### Step 1: Identify the given parameters
- Mass of the ball, \( m = 160 \, \text{g} = 0.16 \, \text{kg} \) (convert grams to kilograms)
- Initial speed, \( u = 10 \, \text{m/s} \)
- Angle of projection, \( \theta = 60^\circ \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the horizontal and vertical components of the initial velocity
- The horizontal component of velocity, \( u_x = u \cos \theta = 10 \cos 60^\circ = 10 \times \frac{1}{2} = 5 \, \text{m/s} \)
- The vertical component of velocity, \( u_y = u \sin \theta = 10 \sin 60^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s} \)

### Step 3: Determine the maximum height reached by the ball
The maximum height \( H \) can be calculated using the formula:
\[ H = \frac{u_y^2}{2g} \]
Substituting the values:
\[ H = \frac{(5\sqrt{3})^2}{2 \times 10} = \frac{75}{20} = \frac{15}{4} \, \text{m} \]

### Step 4: Calculate the angular momentum at the highest point
At the highest point, the vertical component of velocity is zero, and the horizontal component remains constant. The angular momentum \( L \) is given by:
\[ L = m \cdot v \cdot r_{\perpendicular} \]
Where:
- \( v = u_x = 5 \, \text{m/s} \) (horizontal component of velocity)
- \( r_{\perpendicular} = H = \frac{15}{4} \, \text{m} \)

Substituting the values:
\[ L = 0.16 \, \text{kg} \cdot 5 \, \text{m/s} \cdot \frac{15}{4} \, \text{m} \]
Calculating step-by-step:
1. \( 0.16 \cdot 5 = 0.8 \)
2. \( 0.8 \cdot \frac{15}{4} = 0.8 \cdot 3.75 = 3 \, \text{kg m}^2/\text{s} \)

### Final Answer
The angular momentum of the ball at the highest point of the trajectory with respect to the point from which it was thrown is approximately \( 3 \, \text{kg m}^2/\text{s} \).

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