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 $t = \dfrac{T}{\sqrt2}$, is ______ .
[Take $g = 10\ \text{m s}^{-2}$]
Answer: (D) $100\sqrt2\hat i + (100\sqrt2 - 200)\hat j$ N s
Trajectory at $45^\circ$: $y = x - \dfrac{gx^2}{u^2}$.
$$10 = 20 - \frac{10\times400}{u^2} \Rightarrow u^2 = 400 \Rightarrow u = 20\ \text{m s}^{-1}$$
So $u_x = u_y = 10\sqrt2\ \text{m s}^{-1}$ and $T = \dfrac{2u_y}{g} = 2\sqrt2$ s, giving $t = \dfrac{T}{\sqrt2} = 2$ s.
At $t = 2$ s: $v_x = 10\sqrt2$, $v_y = 10\sqrt2 - 20$.
$$\vec p = 10\vec v = 100\sqrt2\,\hat i + (100\sqrt2 - 200)\hat j\ \text{N s}$$
Solution by Sreeraj P, M.Sc Physics