Max damage to target by projectile 
The image above is that of a past exam question. 
Unfortunately I am having trouble deriving a solution, my method is as follows: 
It is clear that the launch velocity for max range ($\theta = 45˚$) is:
$$
V_{max} = {(30g)}^{1/2}
$$
I assumed the landing velocity symmetric to the launch velocity, then applied the coefficient of restitution in the y-plane: 
$$
v_{land} = Vcos\theta i - Vsin\theta j
$$
$$
v_{bounce} = Vcos\theta i + eVsin\theta j
$$
Using the range equation given:
$$
45m = \frac{V^2sin2\theta}{g}+\frac{eV^2sin2\theta}{g}
$$
This gives no meaningful answer for $\theta$ for any value of $v$. What is wrong with my thinking? This is all assuming max velocity is used.  
EDIT
I believe the source of the given expression is as follows:
In the vertical direction:
$$u=Vsin\theta, t = T_{flight}, a = -g, s_y=0$$
Therefore:$$s = ut + \frac{1}{2}at^2$$
$$
T_{flight} = \frac{Vsin\theta}{g}
$$
Now in the horizontal plane the range is:
$$
s=u_xt = Vcos\theta T_{flight} = \frac{\frac{1}{2}V^2sin\theta cos\theta}{g}  = \frac{V^2sin2\theta}{g} 
$$
Why cannot the $V^2$ term not be left as is seeing as it looks like it just expresses the magnitude of the launch velocity?
 A: Derivation of range equation and maximum damage condition
A ball shot at an angle $0\leq\theta\leq \pi/2$ with velocity $\bar{v}\equiv(v_x,v_y)=V (\cos\theta,\sin\theta)$, will reach a (horizontal) distance $d=v_x t$, where $t= 2 v_y / g$ is the time that the ball needs for going up and returning to the ground which is equal to twice the time required to accelerate from (vertical) velocity equal to zero (at maximum height) to maximum velocity $v_y$ at ground level. (Note that if you throw something up it takes the same amount of time to reach the maximum height as it takes to return from maximum height to the start.) Taking these equations together gives:
$$d=\frac{2v_xv_y}{g}$$
After each bounce the vertical velocity is diminished by a factor $e=0.8$.
So the distance after $n$ bounces becomes:
$$d(n,\theta)=\frac{2v_xv_y}{g}(1+e+e^2+...+e^n)=\frac{V^2\sin 2\theta}{g}\frac{1-e^{n+1}}{1-e}$$
Assuming that the launch velocity $V$ is fixed (cannot be adjusted), we have $V^2/g=30\textrm{m}$ from the question (maximum distance without bounce is achieved for $\theta=45˚$)
The question asks for the launch angle at which maximum damage is inflicted at the target. This is equivalent to the ball arriving with maximum kinetic energy. We are thus looking for solutions $n, \theta$ that satisfy:
$$45 \textrm{m}=d(n,\theta)\leftrightarrow \sin 2\theta = \frac{3(1-e)}{2(1-e^{n+1})}=:f(n)\tag{1}$$
$$E_\textrm{kin}\propto v_x^2+e^{2n}v_y^2 \propto \cos^2\theta + e^{2n}\sin^2\theta=:\tilde{E}, \textrm{ is max}\tag{2}$$ 
Two solutions for each $n$, exclude one solution
Equation (1) has for all $n\geq 1$ two solutions $\theta_\pm=\pi/4 \pm \Delta $ with $0<\Delta<\pi/4$, which are symmetrical to $\pi/4$.This represents the fact that to reach a certain distance which is not the maximum distance you can either shoot at angle larger or smaller than 45 degrees.The kinetic energy at impact corresponding to these solutions is:
$$\tilde{E}_\pm = \cos^2(\pi/4\pm \Delta)+e^{2n} \sin^2(\pi/4\pm \Delta)\\ = \frac{1}{2}\left( (\cos\Delta\mp \sin\Delta)^2 + e^{2n} (\cos\Delta\pm \sin\Delta)^2\right)\\
=\frac{1}{2}\left( 1 \mp \sin 2\Delta + e^{2n} (1 \pm  \sin 2\Delta )\right)$$ 
Since $\tilde{E}_--\tilde{E}_+ = \sin 2\Delta (1-e^{2n})>0$, only $\theta_-$ needs to be taken into account when looking for a maximum damage solution.
Solution
With the help of equation (1) the kinetic energy $\tilde{E}_-$ can be written as a function of $n$, the number of bounces only. (It is not a pretty function so it is left out here.) Interestingly, $\tilde{E}_-(n)$ is a monotonically growing function of $n$ which shows that the maximum energy/damage is achieved in the limit $n\to\infty$.In this limit, from equation (1) we have: 
$$\sin 2 \theta_\infty=\frac{3}{10}$$
$$\tilde{E}_- = \cos^2 \theta_\infty$$
Discussion
The solution corresponds to the ball shot at a relatively low angle bouncing off the ground many (infinitely many) times just reaching the target and not going any further.
Why a low angle solution?
If shooting at a low angle, the horizontal component of the velocity, $v_x$ is larger than the vertical component, $v_y$ and correspondingly the initial kinetic energy stems mostly from the horizontal velocity. Since the horizontal velocity does not change during flight or bounces, all of the (large) horizontal velocity contribution to the energy is preserved until impact. Conversely if you shot the ball almost vertically, most of the kinetic energy is in the vertical component of the velocity which does get diminished due to the bounces.
Why a finite, non-zero shooting angle?
In the model used, the ball will not bounce an infinite distance. With each bounce it loses energy and the distance covered between bounces diminishes by  a factor of $e$ from bounce to bounce, making it eventually stop at some maximum distance. Mathematically it is due to the convergence of geometric series. So in this example shooting at a smaller angle than $\theta_\infty$, the ball will not reach the target at 45 m, but stop before.
Why not the $n=1$ solution?
Naively one could think that bouncing the minimum number of times is giving the largest damage because in this case only one or a couple of times the vertical velocity is diminished. However this ignores that the horizontal velocity component also contributes to the impact damage.
Is this practical?
The actual type of target is not known from the question but would have a great influence as real targets would behave differently to projectiles coming from the side or from above. This solution here assumes a point target where horizontal and vertical component of the velocity have the same effect (=enter on equal footing in the kinetic energy/damage). 
An infinite number of bounces, i.e. the limiting solution given above, is not practical for a number of reasons. Realistic (non-perfect) ground might easily introduce a large enough perturbation to make the projectile stop before reaching the target. However, $\tilde{E}_-(n)$, quickly increases at small $n$ from $\tilde{E}_-(1)\approx 0.93$ for one bounce to $\tilde{E}_-(10)\approx 0.973$ for 10 bounces which is already close to the limit of $\tilde{E}_\infty\approx 0.977$ So even if one goes with a slightly larger angle than $\theta_\infty$ and consequently fewer bounces to reach the target, the kinetic energy is almost the same as the (theoretical) limit.
A: If the maximum range is d = 30 m, you can't score a direct hit at 45 m - which is why your initial approach doesn't give you a solution for $\theta$.
However, if you allow a single bounce (and assuming that the ball slips on impact so there is no friction in the horizontal direction), then we can see what happens to the angle after the bounce:

The horizontal velocity doesn't change after the bounce - but the vertical velocity does. 
Now we get the distance traveled by multiplying the horizontal velocity by the total time of flight for the two bounces. We calculate this from the initial vertical velocity and the coefficient of restitution $\eta$. 
Starting from $y=0$ with velocity $v$, you return to the ground in time $t = \frac{2v}{g}$. After the bounce, vertical velocity is reduced by $\eta$. The second bounce will therefore happen at a time $t_1 \left(1+\eta\right)$. Now we need to make sure that this bounce occurs when the ball is at a distance of D = 45 m; in other words,
$$v_h t_1 \left(1+\eta\right) = D$$
Now $v_h = v \cos\theta$ and $v_v = v \sin\theta$, so putting it all together we get
$$\frac{2v\sin\theta}{g}v\cos\theta\left(1+\eta\right)=D$$
But we also know the original velocity $v$ from the first experiment, namely $v^2 = d g$. Putting that into the above equation, you will find the relationship you need.
The interesting question was raised whether it would be better (in terms of the final velocity) to have more bounces. In principle, a "skimming shot" would give you less loss of energy per bounce, but more bounces. It's reasonable to ask what would be better.
One can take two approaches to get some insight into this. The first way involves calculating the final range of a projectile as a function of launch angle, assuming it will "bounce to rest". In that case, the total time of flight is given by
$$\begin{align}t &= \frac{2v\sin\theta}{g}\left(1+\eta+\eta^2+...\right) \\
&= \frac{2v\sin\theta}{g}\left(\frac{1}{1-\eta}\right)\end{align}$$
The range is then given by 
$$ D = t~ v\cos\theta = \frac{v\sin 2\theta }{g(1-\eta)}$$
This shows the range is extended by the bouncing- but 45 degrees would still be the optimal angle if your goal is "distance".
More interesting (and relevant) is the calculation of the final velocity after $n$ bounces, given a known initial velocity and final (target) range. Then we find the angle needed by solving
$$D = \frac{V^2 \sin 2\theta}{g}\left(\frac{1-\eta^{n+1}}{1-\eta}\right)$$
And the impact velocity follows from
$$V_i = \sqrt{v_h^2+v_v^2}\\
= V\sqrt{\cos^2\theta +(\eta\sin\theta)^{2\eta}}$$
This is an ugly function; solving for different values of $n$, you get the interesting result that the impact velocity initially gets lower, then higher again as the number of bounces increases - and this is a function of the desired range and the coefficient of restitution. For the values given, you can compute the final velocity after 1, 2, ...n bounces - given you are trying to reach the same target distance. The trajectories you get look like this:

While the final velocity looks like this:

This shows that the trajectory that "just" hits the target after the largest number of bounces is actually optimal.
A: Your 1st equation (just before your EDIT) is correct, it must be your arithmetic which is going wrong. There is a solution for 1 bounce.

The formula for range can be written in terms of the horizontal and vertical components of velocity $v_x, x_y$ as
$R=\frac{2v_x v_y}{g}$.
After each bounce $v_y$ is reduced to $ev_y$ while $v_x$ is unchanged. So after $n$ bounces the range will be
$R(n)=\frac{2v_xv_y}{g}(1+e+e^2+...+e^n)=(1+e+e^2+...+e^n)R_0 \sin2\theta$
where $R_0=\frac{v^2}{g}=30m$ is the maximum range without bouncing $(\theta=45^{\circ})$.
To reach a range of $45m$ the superball must bounce at least once. Maximum damage to the target requires maximum KE, which is accomplished by maximising the final vertical component of velocity $e^n v_y$ (since $v_x$ is constant), which requires minimising the number of bounces $n$ (because $e \lt 1$). 
For $n=1$ bounce there is a solution for $\theta$.
