Assume no friction and uniform gravity g.
If you throw a stone at point (0, h), with velocity (v cos θ, v sin θ), then we get
\begin{align}
d &= vt\cos\theta && (1) \\
0 &= h + vt\sin\theta - \frac12 gt^2 && (2)
\end{align}
The only unknown to be solved is t (total travel time). We could eliminate it by using $t = \frac d{v\cos\theta}$ to get
$$ 0 = h + d\tan\theta - \frac{gd^2\sec^2\theta}{2v^2}\qquad(3) $$
Then we compute the total derivative with respect to θ:
\begin{align}
0 &= \frac d{d\theta}\left(d\tan\theta\right) - \frac g{2v^2}\frac d{d\theta}\left(d^2\sec^2\theta\right) \\
&= \ldots
\end{align}
and then set $\frac{dd}{d\theta}=0$ (because it is maximum) to solve d:
$$ d = \frac{v^2}{g\tan\theta} $$
Substitute this back to (3) gives:
\begin{align}
h &= \frac{v^2}g \left( \frac1{2\sin^2\theta} - 1\right) \\
\Rightarrow \sin\theta &= \left( 2 \left(\frac{gh}{v^2} + 1\right) \right)^{-1/2}
\end{align}
This is the closed form of θ in terms of h.
Looking at the vertical forces on the crate, you must have an equation like $T \sin 30^{\circ} + F = mg$, where $T$ is the tension on the rope and $F$ is the normal reaction from the floor. This gives $F = 15\times 10 - 69 \sin 30^{\circ} = 115.5\,$N.
Now looking at the horizontal forces, you have $T \cos 30^{\circ} \approx 60$ in one direction and frictional force of $\mu F = 0.4 \times 115.5 \approx 46 $ in the other direction. So, there is a net unbalanced force of about $60-46 = 14\,$N which will accelerate the crate at $\frac{14}{15}$ $m/s^2$.
Total work done is then work done against friction $+$ kinetic energy gained by the acceleration, but this will be equal to force in the horizontal direction times displacement anyway, so $60*10 = 600\,$J.
Best Answer
I am surprised no one seems to have answered this.
I get $$\pi/4 + \alpha/2$$.
If the line is given by $y = x \tan \alpha$, with $\alpha$ acute and we throw from the origin at an angle $\theta$ from the x-axis, at velocity $1$, then we have that the projectile satisfies, assuming gravity $g=2$ (in appropriate units)
$\displaystyle y = t\sin\theta - t^2$, $\displaystyle x = t\cos \theta$
The time at which it intersects the line again is given by
$\displaystyle t\sin\theta - t^2 = t \tan\alpha \cos \theta$ and so
$\displaystyle t = \sin \theta - \tan \alpha \cos \theta$
It is enough to maximize the horizontal distance travelled by the projectile, which is given by
$\displaystyle \cos \theta (\sin \theta - \tan \alpha \cos \theta)$
With little manipulation, we need to maximize
$\displaystyle \sin(2\theta - \alpha)$
which gives the result.