In the projectile motion shown in figure, it is given that $t_{AB}=2s$ and the particle is projected at t = 0, then which of the following option(s) is/are correct (Consider g = 10 $m{s}^{-2}$)

The correct options are
A Particle is at point B at 3 s
B Maximum height of projectile is 20 m
C Initial vertical component of velocity is 20 $m{s}^{-1}$
D Horizontal component of velocity is 20 $m{s}^{-1}$
Horizontal component of velocity remains unchanged
$X_{QA}=20m=\frac{X_{AB}}{2}\therefore t_{QA}=\frac{t_{AB}}{2}=1s$
For ‘AB’ part of projectile motion
$T=2s=\frac{2u_y}{g}\therefore u_y=10m/sH=\frac{u_y^2}{2g}=\frac{\left({10}^2\right)}{2\times 10}=5m$
$\therefore$ Maximum height of the projectile = 15 + 5 = 20 m
$t_{OB}=t_{OA}+t_{AB}=1+2=3s$
For complete projectile motion,
$T=2\left(t_{OA}\right)+t_{AB}=4s=\frac{2u_y}{g}\therefore u_y=20m/s{u}_x=\frac{AB}{t_{AB}}=\frac{40}{2}=20m/s$