Kinematics - Two particles release from same height ...

$20 = 8.18t + \frac{1}{2}(\frac{12}{20}g)t^2$

$t = 1.564 ~ \text{sec}$       (okay)
 

Components of 8.18 m/s
$v_{ox} = \frac{4}{5}(8.18) = 6.544 ~ \text{m/sec}$

$v_{oy} = \frac{3}{5}(8.18) = 4.908 ~ \text{m/sec}$
 

Check the time in horizontal motion (a = 0 and s = 16 m)
$s = v_ot + \frac{1}{2}at^2$

$16 = 6.544t + 0$

$t = 2.445 ~ \text{sec}$       (not okay)
 

Check the time in vertical motion (a = g and s = 12 m)
$s = v_ot + \frac{1}{2}at^2$

$12 = 4.908t + \frac{1}{2}gt^2$

$t = 1.142 ~ \text{sec}$       (not okay)

$v_{2y} = v_{oy} + at$

$v_{2y} = 0 + g\sqrt{24/g}$

$v_{2y} = 15.344 ~ \text{m/sec}$
 

Inclined Velocity
$\dfrac{v_2}{15.344} = \dfrac{20}{12}$

$v_2 = 25.573 ~ \text{m/sec}$
 

Horizontal Velocity
$\dfrac{v_{2x}}{15.344} = \dfrac{16}{12}$

$v_{2x} = 20.459 ~ \text{m/sec}$
 

Check the Inclined Motion
$v^2 = {v_o}^2 + 2as$

${v_2}^2 = {v_o}^2 + 2aL$

$25.573^2 = 0^2 + 2a(20)$

$a = 16.349 ~ \text{m/sec}^2$
 

$v = v_o + at$

$v_2 = v_o + at$

$25.573 = 0 + 16.349t$

$t = 1.564 ~ \text{sec}$       (okay)
 

Check the Horizontal Motion
$v^2 = {v_o}^2 + 2as$

${v_{2x}}^2 = {v_{ox}}^2 + 2a_x L_x$

$20.459^2 = 0^2 + 2a_x (16)$

$a_x = 13.08 ~ \text{m/sec}^2$
 

$v = v_o + at$

$v_{2x} = v_{ox} + a_x t$

$20.459 = 0 + 13.08t$

$t = 1.564 ~ \text{sec}$       (okay)
 

Additional checking: the a_y must be equal to g
${a_x}^2 + {a_y}^2 = a^2$

$13.08^2 + {a_y}^2 = 16.349^2$

$a_y = 9.81 ~ \text{m/sec}^2$       (okay)