Ans : Initial velocity of the 1st particle $u_1$=20 m /sec assume acceleration due to gravity g = 10m/$sec^2$.
Maximum height traveled by this particle $H_{max}$= $\frac{u_1^2}{2g}$ = $\frac{20^2}{2\times10}$.
$H_{max}$= $\frac{400}{20}$ ; $H_{max}$= 20m.
Time of ascent $t_a$= u/g =20/10 =2 sec.