Is there a better, faster way to do this projectile motion question? The question is

In a combat exercise, a mortar at M is required to hit a target at O, which is taking cover 25 m behind a structure of negligible width 10 m tall. This mortar can only fire at an angle of 45 degrees to the horizontal, but can fire shells of any velocity. Find the minimum initial velocity required to hit the target.
  

I solved it as follows.
Forming a parabola with $ r_v $ against $ r_h $:
$ r_v = u_vt + \frac{1}{2}a_vt^{2} $ (1) 
$ r_h = u_ht + \frac{1}{2}a_ht^{2} $
but $ a_h = 0 $
so $ r_h = u_ht $
$ t = \frac{r_h}{u_h} $ (2)
Substituting (2) into (1):
$ r_v = u_v\frac{r_h}{u_h} + \frac{1}{2}a_v\frac{r_h^{2}}{u_h^{2}} $ (3) 
and because the angle of inclination is 45°
$ u_v = u_h = \frac{u}{\sqrt{2}} $ (4)
From (4) and (3):
$ r_v = r_h^{2}\frac{a}{u^{2}} + r_h $ (5) 
Let the distance between the mortar and the building be $ d $.
Then when $ r_h = d + 25 $, $ r_v = 0 $. (6)
From (5) and (6):
$ 0 = (d + 25)\frac{a}{u^{2}} + 1 $ 
so $ u^2 = -a(d + 25) $ (7)
Substituting (7) into (5):
$ r_v = -r_h^{2}\frac{1}{(d + 25)} + r_h $ (8) 
We also know that to clear the building, when $ r_h = d $, $ r_v > 10 $. (9)
From (9) and (8):
$ 10 < -d^{2} \frac{1}{(d + 25)} + d $
After simplifying... ($ d + 25 $ is positive)
$ d > \frac{50}{3} $ (10)
Rearranging (7):
$ d = -\frac{u^{2}}{a} - 25 $ (11)
And then from (10) and (11) and with $ a = -9.8 $:
$ \frac{50}{3} < \frac{u^{2}}{9.8} - 25 $
Simplifying, and with the knowledge that $ u > 0 $:
$ u > 20.2073... $
So the minimum initial velocity required to hit the target is 20 m/s (2 s. f.).
Huzzah!
My question is: is there a faster way to solve the problem?
 A: Whether this is any faster is debatable, but you could do it this way:
The trajectory of the shell is symmetric, so $M$ firing a shell at $O$ is the same as $O$ firing a shell at $M$. So all you have to do is consider $O$ firing the mortar at 45° and ask what is the minimum velocity required to clear the wall. So, if $O$ fires the shell at 45° the equations of motion are:
$$\begin{align}
x &= t \frac{v}{\sqrt{2}} \\
y &= t \frac{v}{\sqrt{2}} - \frac{1}{2} g t^2
\end{align}$$
If we require that the trajectory passes through the point $(25, 10)$ then this gives us two simultaneous equations in $t$ and $v$:
$$\begin{align}
25 &= t \frac{v}{\sqrt{2}} \\
10 &= t \frac{v}{\sqrt{2}} - \frac{1}{2} g t^2
\end{align}$$
It's $v$ we're interested in, so we rearrange the first equation to get:
$$ t = \frac{25\sqrt{2}}{v} $$
and substitute it in the second equation to get:
$$ 10 = \frac{25\sqrt{2}}{v} \frac{v}{\sqrt{2}} - \frac{1}{2} g \left( \frac{25\sqrt{2}}{v} \right)^2 $$
and a quick rearrangement gives:
$$ v = \sqrt{\frac{g \space 25^2}{15}} = 20.2 m/s $$
