A particle is projected at an angle of ${45}^{\mathrm{o}}$ . After 1 sec, it breaks into two equal parts. One part stops and the other part attains the height of 20 m after the breaking of the particle. Find the velocity of projection ( g = 10 $\mathrm{m}/{\mathrm{s}}^2$ )

$\frac{{\mathrm{V}}_{\mathrm{y}}^2}{2\mathrm{g}}=20\mathrm{m}$

$\Rightarrow {\mathrm{V}}_{\mathrm{y}}=20\mathrm{m}/\mathrm{s}$ = vertical velocity of the second part.

From conservation of momentum  $\mathrm{m}\times 0+{\mathrm{mV}}_{\mathrm{y}}=\left(2\mathrm{m}\right){\mathrm{V}}_{{\mathrm{y}}_{\mathrm{o}}}$
${\mathrm{V}}_{{\mathrm{y}}_0}$  vertical velocity of particle after one second before breaking of it
= $\frac{{\mathrm{V}}_{\mathrm{y}}}{2}=\frac{20}{2}=10\mathrm{m}/\mathrm{s}$

But  $V_{y_0}=10=u_y-gt$
 $\Rightarrow {\mathrm{u}}_{\mathrm{y}}=10+10\times 1=20\mathrm{m}/\mathrm{s}$
Now, $u_y=u\sin {45}^o$

$\Rightarrow \mathrm{u}=\frac{{\mathrm{u}}_{\mathrm{y}}}{\sin {45}^{\mathrm{o}}}=20\sqrt{2}\mathrm{m}/\mathrm{s}$