Kinematics is fun!!!

Here is another method to solve.

Let's say the stone being thrown upward as $I$ and the downward one as $II$ . Those stones will meet in the air at the same time, let's call it $t$ .

Assume the $x$ as the point where stones meet, the distances the $I$ stone had been travelled to $x$ as $S_{I}$ , and the distances the $II$ stone had been travelled to $x$ as $S_{II}$ .

It will appears that $\boxed{S_I + S_{II} = h}$

$S_I = v_{I}.t_{I} - \frac{1}{2}gt_{I}^2$ .

But be carefull, the $II$ stone being thrown $2$ second after the $I$ stone. That means the $I$ stones travelled $2$ seconds longer than the $II$ stone, or : $t_{I} = t + 2$ . Then :

$S_I = 50(t + 2) - \frac{1}{2}10(t + 2)^2$

$S_I = 30t - 5t^2 + 80$

For the $II$ stone, the time is the same, $t_{II} = t$ .

$S_{II} = v_{II}t + \frac{1}{2}gt^2$

$S_{II} = 3t + \frac{1}{2}10t^2$

$S_{II} = 3t + 5t^2$

Back to the main equation :

$h = S_{I} + S_{2}$

$278 = (30t - 5t^2 + 80) + (3t + 5t^2)$

$198 = 33t => \boxed{t = 6}$

The height of the point $x$ will be the same as $S_{I}$

$x = S_I = 30t - 5t^2 + 80$

$x = 30.6 - 5.6^2 + 80 = \boxed{80 m}$