Not 45.0

With the $x$ -axis horizontal and the origin at ground level below the point of projection, the trajectory of the ball has equation $x \; = \; 10t\cos\phi \qquad \qquad y \; = \; 10 + 10t\sin\phi - 5t^2$ if it is projected at an angle of $\phi$ above the horizontal. The ball hits the ground when $y=0$ , so when $t^2 - 2t\sin\phi - 2 = 0$ , which happens when (for future time) $t = \sin\phi + \sqrt{\sin^2\phi + 2}$ . Thus the horizontal range of the ball is $R(\phi) \; = \; 10\big[\sin\phi\cos\phi + \cos\phi\sqrt{\sin^2\phi + 2}\big]$ To find the maximum range, find the turning point of $R$ . Since $R'(\phi) \; = \; \frac{10\big[\cos2\phi\big(\sqrt{\sin^2\phi + 2} + \sin\phi\big) - 2\sin\phi\big]}{\sqrt{\sin^2\phi + 2}}$ we can see that $R'(\tfrac16\pi) =0$ , and this angle of projection can easily be shown to give the maximum range. Thus the answer is $\boxed{30}$ .