1. Given data

• Mass of particle: $m = 100\text{ g} = 0.1\text{ kg}$
• Initial speed: $u = 20\text{ m/s}$
• Projection angle: $45^\circ$
• Time: $t = 2\text{ s}$
• Acceleration due to gravity: $g = 10\text{ m/s}^2$

We need the magnitude of angular momentum about the starting point at $t=2\text{ s}$.

2. Resolve initial velocity into components

Since the particle is projected at $45^\circ$,

$u_x = u\cos 45^\circ = 20\cdot \frac{1}{\sqrt{2}} = 10\sqrt{2}$ $u_y = u\sin 45^\circ = 20\cdot \frac{1}{\sqrt{2}} = 10\sqrt{2}$

3. Position of the particle at $t=2$ s

Horizontal position:

$x = u_x t = 10\sqrt{2}\cdot 2 = 20\sqrt{2}$

Vertical position:

$y = u_y t - \frac{1}{2}gt^2$ $y = 10\sqrt{2}\cdot 2 - \frac{1}{2}\cdot 10 \cdot (2)^2$ $y = 20\sqrt{2} - 20$

So,

$\vec r = (20\sqrt{2})\hat i + (20\sqrt{2}-20)\hat j$

4. Velocity of the particle at $t=2$ s

Horizontal velocity remains constant:

$v_x = 10\sqrt{2}$

Vertical velocity:

$v_y = u_y - gt = 10\sqrt{2} - 10\cdot 2 = 10\sqrt{2} - 20$

So,

$\vec v = (10\sqrt{2})\hat i + (10\sqrt{2}-20)\hat j$

Momentum:

$\vec p = m\vec v = 0.1\vec v$

5. Angular momentum about the starting point

Angular momentum is

$\vec L = \vec r \times \vec p = m(\vec r \times \vec v)$

In 2D, magnitude is

$L = m|xv_y - yv_x|$

Substitute values:

$xv_y = (20\sqrt{2})(10\sqrt{2}-20)$ $= 20\sqrt{2}\cdot 10\sqrt{2} - 20\sqrt{2}\cdot 20$ $= 400 - 400\sqrt{2}$

Now,

$yv_x = (20\sqrt{2}-20)(10\sqrt{2})$ $= 20\sqrt{2}\cdot 10\sqrt{2} - 20\cdot 10\sqrt{2}$ $= 400 - 200\sqrt{2}$

Therefore,

$xv_y - yv_x = (400 - 400\sqrt{2}) - (400 - 200\sqrt{2})$ $= -200\sqrt{2}$

Hence,

$|xv_y - yv_x| = 200\sqrt{2}$

So,

$L = m\cdot 200\sqrt{2} = 0.1\cdot 200\sqrt{2} = 20\sqrt{2}$

Thus,

$L = \sqrt{800}\;\text{kg m}^2\!/\text{s}$

So,

$K = 800$

6. Comparison with stored answer

Derived answer: $800$

Stored correct answer: $800$

They match.