A body of mass 10   kg is projected at an angle of 45 ° with the horizontal. The trajectory of the…

It is given that the projectile passes through $\left(20,10\right)$ and angle of projection is $\theta =45°$. Using the trajectory equation,

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

$\Rightarrow 10=20-\frac{\left(10\right)\left(100\right)}{u^2}$

$\Rightarrow u=20 \mathrm{m} {\mathrm{s}}^{-1}$

Now the time of flight of projectile,

$T=\frac{2u\sin \theta }{g}$

$$\begin{gathered} \Rightarrow T=\frac{\left(2\right)\left(20\right)}{\sqrt{2}\left(10\right)}=2\sqrt{2} \mathrm{s} \\ \Rightarrow \frac{T}{\sqrt{2}}=2 \mathrm{s} \end{gathered}$$

Now velocity of the projectile at any time $t$ can be written as,

$\overrightarrow{v}=u\cos \theta \hat{\mathrm{i}}+\left(u\sin \theta -gt\right)\hat{\mathrm{j}}$

$\Rightarrow \overrightarrow{v}=10\sqrt{2}\hat{\mathrm{i}}+(10\sqrt{2}-10\left(2\right)]\hat{\mathrm{j}} \mathrm{m} {\mathrm{s}}^{-1}$

Therefore, momentum $\overrightarrow{p}=M\overrightarrow{v}=100\sqrt{2}\hat{\mathrm{i}}+\left(100\sqrt{2}-200\right)\hat{\mathrm{j}} \mathrm{N} \mathrm{s}$