1. Write the projectile equation

For projection at angle $45^\circ$ with speed $u$:

$u_x=u\cos45^\circ=\frac{u}{\sqrt2},\qquad u_y=u\sin45^\circ=\frac{u}{\sqrt2}$

Given that the trajectory passes through $(x,y)=(20,10)$.

The trajectory equation is

$y=x\tan\theta-\frac{gx^2}{2u^2\cos^2\theta}$

With $\theta=45^\circ$, we have $\tan45^\circ=1$ and $\cos^245^\circ=\frac12$. Hence

$y=x-\frac{gx^2}{u^2}$

Substitute $x=20$, $y=10$, $g=10$:

$10=20-\frac{10\cdot 20^2}{u^2}$

$10=20-\frac{4000}{u^2}$

$\frac{4000}{u^2}=10$

$u^2=400 \Rightarrow u=20\,\text{m/s}$

2. Find components of initial velocity

$u_x=u_y=\frac{20}{\sqrt2}=10\sqrt2\,\text{m/s}$

3. Find time of flight

For projectile motion,

$T=\frac{2u\sin45^\circ}{g}$

$T=\frac{2\cdot 20\cdot \frac{1}{\sqrt2}}{10}=2\sqrt2\,\text{s}$

Therefore,

$t=\frac{T}{\sqrt2}=\frac{2\sqrt2}{\sqrt2}=2\,\text{s}$

4. Velocity at $t=2$ s

Horizontal velocity remains constant:

$v_x=u_x=10\sqrt2$

Vertical velocity:

$v_y=u_y-gt=10\sqrt2-10(2)=10\sqrt2-20$

So,

$\vec v=10\sqrt2\,\hat i+(10\sqrt2-20)\,\hat j$

5. Momentum vector

Given mass $m=10$ kg,

$\vec p=m\vec v$

$\vec p=10\left[10\sqrt2\,\hat i+(10\sqrt2-20)\,\hat j\right]$

$\boxed{\vec p=100\sqrt2\,\hat i+(100\sqrt2-200)\,\hat j}$

6. Compare with options

This matches:

$\boxed{\text{Option D}}$

7. Verification with stored answer

Stored correct answer: $D$

Derived answer: $D$

So they agree.