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 the highest point of the trajectory with respect to the point from which the ball is thrown is nearly `(g=10ms^(-2))`

Options

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

Answer

To find the angular momentum of the ball at the highest point of its trajectory, we can follow these steps:

### Step 1: Identify the given data
- Mass of the ball, \( m = 160 \, \text{g} = 0.16 \, \text{kg} \) (conversion from 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 the velocity \( v_x = u \cos \theta = 10 \cos 60^\circ = 10 \times \frac{1}{2} = 5 \, \text{m/s} \)
- The vertical component of the velocity \( v_y = u \sin \theta = 10 \sin 60^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s} \)

### Step 3: Determine the maximum height reached by the ball
At the highest point, the vertical velocity becomes zero. We can use the following kinematic equation to find the maximum height \( h \):

\[ v_y^2 = u_y^2 - 2gh \]
Setting \( v_y = 0 \) at the highest point:

\[ 0 = (5\sqrt{3})^2 - 2gh \]
\[ 0 = 75 - 20h \]
\[ 20h = 75 \implies h = \frac{75}{20} = 3.75 \, \text{m} \]

### Step 4: Calculate the angular momentum at the highest point
The angular momentum \( L \) about the point of projection is given by:

\[ L = m \cdot v_x \cdot r_{\perp} \]

Where:
- \( r_{\perp} \) is the perpendicular distance from the line of action of the velocity to the point of projection. At the highest point, this distance is equal to the maximum height \( h \).

Thus,

\[ L = m \cdot v_x \cdot h \]
Substituting the known values:

\[ L = 0.16 \, \text{kg} \cdot 5 \, \text{m/s} \cdot 3.75 \, \text{m} \]
\[ L = 0.16 \cdot 5 \cdot 3.75 = 3 \, \text{kg m}^2/\text{s} \]

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

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