Fire up the hill

Suppose that you and your potato gun are at the origin of a standard $xy$ -grid. The base of the hill is then at $(30,0),$ (measured in meters), and the target is at

$(X,Y) = (30 + 80\cos(\frac{\pi}{6}), 80\sin(\frac{\pi}{6})) = (30 + 40\sqrt{3}, 40).$

The equations of motion of the potato projectile are then

$x = v_{0}\cos(\phi)t$ and $y = v_{0}\sin(\phi)t - \dfrac{g}{2}t^{2},$

where $t = 0$ s at the time of launch and $\phi$ is the launch angle. Substituting $t = \dfrac{x}{v_{0}\cos(\phi)}$ into the equation for $y$ and simplifying yields

$y = x\tan(\phi) - \dfrac{gx^{2}}{2v_{0}^{2}}(\tan^{2}(\phi) + 1),$

which when rewritten as a quadratic in $\tan(\phi)$ becomes

$\dfrac{gx^{2}}{2v_{0}^{2}}\tan^{2}(\phi) - x\tan(\phi) + \left(\dfrac{gx^{2}}{2v_{0}^{2}} + y\right) = 0$

$\Longrightarrow \tan(\phi) = \dfrac{x \pm \sqrt{x^{2} - 4\dfrac{gx^{2}}{2v_{0}^{2}}\left(\dfrac{gx^{2}}{2v_{0}^{2}} + y\right)}}{\dfrac{gx^{2}}{v_{0}^{2}}}.$

Plugging in $x = X = 30 + 40\sqrt{3} m, y = Y = 40 m, g = 9.81 \frac{m}{s^{2}}$ and $v_{0} = 40 \frac{m}{s}$ gives us that

$\tan(\phi) = \dfrac{99.282 \pm 37.176}{60.3736},$

and since we want to find the minimum possible launch angle we choose the $-$ sign, yielding $\tan(\phi_{min}) = 1.0287 \Longrightarrow \phi_{min} = \boxed{0.7995}$ radians.

Adjusted from most efficient of 54.4810315807003 degrees down to 45.8579456319378 degrees such that 0.800372139477866 radians. If a tall height of 1.5 m is considered in, then 44.7448867282823 degrees! Substitutions of time factors to eliminate time for equation of Sine and Cosine of angle of elevation, numerical substitution shall be easiest to obtain the angle without sorting out Cosine value or Sine value; not solving the equation really. Every item can be checked using Excel to confirm for logical correctness. 66.0861970556276 degrees is another greater one which shall take much more time, changed from 3.56391341736003 s to 6.12305020909730 s of higher peak.

Answer: 0.800372139477866