A particle of mass 2m is projected at an angle of ${45}^0$  with horizontal with a velocity of $20\sqrt{2}m/s$ . After 1 s explosion takes place and the particle is broken into two equal pieces. As a result of explosion one part comes to rest. The maximum height from the ground attained by the other part is $\left(g=10m/s^2\right)$

Horizontal and vertical components of initial velocity are:

$u_x=20\sqrt{2}\cos {45}^0=20m/s$        And $u_y=20\sqrt{2}\sin {45}^0=20m/s$

After 1 s horizontal component remains unchanged while the vertical component becomes     $v_y=u_y-gt$ $=20-\left(10\right)\left(1\right)$        

$=10m/s$ Due to explosion one part comes to rest. Hence, from conservation of linear momentum vertical component of second part will become $v{\text{'}}_y=20m/s$ .

after 1 second  m $\left(20 \overrightarrow{i}+10 \overrightarrow{j}\right)=\frac{m}{2}x 0+\frac{m}{2}\overrightarrow{v}$  $\Rightarrow \overrightarrow{v}=40\overrightarrow{i}+20\overrightarrow{j}$  

 Therefore, maximum height attained by the second part will be       $H=h_1+h_2$

$h_2=\frac{20\times 20}{2\times 10}=20 m$

Here, $h_1=$  height attained in 1 s $=\left(20\right)\left(1\right)-\frac{1}{2}\left(10\right){\left(1\right)}^2$

And $h_2=$  height attained after 1 s       

$=\frac{v{\text{'}}_y^2}{2g}=\frac{{\left(20\right)}^2}{2\times 10}$ $=20m$ $\therefore H=20+15$ $=35m$