Projectile to clear a hemispherical mound I have only just stumbled across this site a few days ago while searching for other things.  This is a great resource.  I have a question for nearly 30 years now when me and my friend were trying to solve the following question from the book "An introduction to the theory of mechanics" by K.E. Bullen.  I just fetched from the library a 7th edition copy and there on p.157, Question 19, it wrote:
A particle is projected from the highest point of a hemispherical mound of radius a.  Prove that it cannot clear the mound unless its initial speed exceeds sqrt(ag/2).
My friend and I worked out a very simple solution which got us the "answer" sqrt(ag/2).  However, we then realized we were definitely wrong.  But we didn't go on finding the correct one as it appeared to be too difficult for us.  Please solve it for me.
(It's my first question.  I may have made mistakes on tags, manner, etiquette ...  please go easy on me.)
 A: The speed of projection is minimal when the projectile just grazes the mound, or equivalently when the projectile fired tangentially to the mound in a vertical plane hits the apex of the mound. Taking the projection point as $(-a\sin\theta,a\cos\theta)$, with the $y$ axis vertical through the centre of the mound, the trajectory in this plane is given by $$y-a\cos\theta=(x+a\sin\theta)\tan\theta-\frac{g(x+a\sin\theta)^2}{2v^2\cos^2\theta},$$where $g$ is the acceleration due to gravity and $v$ is the speed of projection. Substituting in $x=0$ and $y=a$ (the apex) gives, after simplification, $$\frac{2v^2}{ga}=1+\sec\theta.$$ The speed $u$ at the apex is given by $v^2=u^2+2ga(1-\cos\theta)$, or$$u^2=\frac{ga}2(4\cos\theta+\sec\theta-3).$$To minimize this, differentiate with respect to $\theta$ and equate to zero, giving$$-4\sin\theta+\sec\theta\tan\theta=0,$$or $\theta=\frac13\pi.$ Finally, this gives $u^2=\frac12ga$ or $u=\sqrt{\frac12ga}$.
A: If the launch is lateral, the problem is fairly easy.
Let's have $y(t)$ the height of the projectile and $x(t)$ its lateral position, with $(0,0)$ being the center of the hemishparical mound of radius $a$.
Then you have $y(t)=-\frac 12 gt^2 + y(0)$ and $x(t)=v_0t$ (with $y(0)=a$) 
Then the projectile hits the ground at $t=t_1$ when $y(t_1)=0$, that is
$-\frac 12 gt_1^2 + a=0$ or $t_1=\sqrt{\dfrac{2a}{g}}$
Then $x(t_1)=v_0\sqrt{\dfrac{2a}{g}}$
But we want $x(t_1) \geq a$ So the limit case is
$v_0\sqrt{\dfrac{2a}{g}}=a$ or $v_0=\sqrt{\dfrac{ag}{2}}$
