A body of mass 10 kg is projected at an angle of $$45^\circ$$ with the horizontal. The trajectory of the body is observed to pass through a point (20, 10). If $$T$$ is the time of flight, then its momentum vector, at time $$t = \frac{T}{\sqrt{2}}$$, is _____ . [Take $$g = 10 \ m s^{-2}$$]

A body of mass 10 kg is projected at $$45^\circ$$ with the horizontal, and its trajectory passes through the point (20, 10). We need to determine the momentum vector at time $$t = \frac{T}{\sqrt{2}}$$, where $$T$$ is the total time of flight.

The trajectory equation for projectile motion is $$y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}.$$ With $$\theta = 45^\circ$$, $$\tan45^\circ = 1$$ and $$\cos^245^\circ = \tfrac12$$, this becomes $$y = x - \frac{gx^2}{u^2}.$$ Substituting the point $$(20,10)$$ and $$g = 10$$ yields $$10 = 20 - \frac{10\times400}{u^2},$$ so $$\tfrac{4000}{u^2} = 10$$ and hence $$u^2 = 400$$, giving $$u = 20\text{ m/s}.$$ The horizontal and vertical components of the initial velocity are $$u_x = u\cos45^\circ = 20\times\frac{1}{\sqrt2} = 10\sqrt2\text{ m/s},\quad u_y = u\sin45^\circ = 10\sqrt2\text{ m/s}.$$ The total time of flight is $$T = \frac{2u\sin\theta}{g} = \frac{2\times20\times\frac{1}{\sqrt2}}{10} = 2\sqrt2\text{ s},$$ so $$t = \frac{T}{\sqrt2} = 2\text{ s}.$$

Because the horizontal velocity remains constant, $$v_x = u_x = 10\sqrt2\text{ m/s},$$ and the vertical velocity at $$t=2$$ s is $$v_y = u_y - gt = 10\sqrt2 - 10\times2 = 10\sqrt2 - 20\text{ m/s}.$$ Thus the momentum vector is $$p = mv = 10(10√2 i + (10√2 - 20) j) = 100√2 i + (100√2 - 200) j N·s$$

Answer: Option D: $$100\sqrt{2}\hat{i} + (100\sqrt{2} - 200)\hat{j}$$ N s