Two particles released from the same height and reached the base at the same time

$12 = \frac{1}{2}gt^2$

$t = \sqrt{\dfrac{24}{g}}$

$v_{Ai-y} = 0.6v_{Ai}$
 

Apply the formula s = v_ot + 0.5at² to the vertical movement of A
$s = v_ot + \frac{1}{2}at^2$

$h = v_{Ai-y}\,t + \frac{1}{2}gt^2$

$12 = (0.6v_{Ai})\sqrt{\dfrac{24}{g}} + \dfrac{g}{2}\left( \sqrt{\dfrac{24}{g}} \right)^2$

$12 = 0.6v_{Ai} \, \sqrt{\dfrac{24}{g}} + \dfrac{g}{2}\left( \dfrac{24}{g} \right)$

$12 = 0.6v_{Ai} \, \sqrt{\dfrac{24}{g}} + 12$

$v_{Ai} = 0$           answer