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

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

$y = xtan\theta - \dfrac{g{x}^{2}}{2{u}^{2}\cos^{2}\theta}$

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

$⇒ u = 20m{s}^{- 1}$

Now the time of flight of projectile,

$T = \dfrac{2usin\theta}{g}$

$⇒ T = \dfrac{\left(2\right)\left(20\right)}{\sqrt{2}\left(10\right)} = 2\sqrt{2}s ⇒ \dfrac{T}{\sqrt{2}} = 2s$

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

\overset{\to}{v} = ucos\theta\overset{^}{i} + \left(usin\theta - gt\right)\overset{^}{j}

⇒ \overset{\to}{v} = 10\sqrt{2}\overset{^}{i} + ( 10\sqrt{2} - 10\left(2\right) ] \overset{^}{j}m{s}^{- 1}

Therefore, momentum \overset{\to}{p} = M\overset{\to}{v} = 100\sqrt{2}\overset{^}{i} + \left(100\sqrt{2} - 200\right)\overset{^}{j}Ns