The problem is from Morin's blue book. He tells to not solve quantitatively from scratch, but I don't really see the point in not doing so, thus I went ahead and tried to do it. This question isn't really about physics but rather what I did wrong in my proof.
A wall has height $h$ and is a distance $\ell$ away. You wish to throw a ball over the wall with a trajectory such that the ball barely clears the wall at the top of its parabolic motion. What initial speed is required?
The given answer is $\sqrt{2gh+g\ell^2/2h}$, but my answer of $\sqrt{2gh+2g\ell^2/h}$ is kind of close but a little bit off. My derivation is this:
We know that $y_{max} = h = \frac{v_o^2sin^2\theta}{2g}$, so $v_o = \sqrt{\frac{2gh}{sin^2\theta}}$, where $v_o$ is the intial speed, and $\theta$ is this angle by which the ball is thrown, so $\theta = \arctan({\frac{h}{\ell}})$. It's well-known that $\sin(\arctan(x))=\frac{x}{\sqrt{1+x^2}}$, so $$\sin^2(\arctan(\frac{h}{\ell}))=\frac{h^2/\ell^2}{1+h^2/\ell^2}.$$
So, we have $$\frac{2gh}{sin^2\theta}=\frac{2gh}{\frac{h^2}{\ell^2}/(1+\frac{h^2}{\ell^2})}=\frac{2gh+2gh^3/\ell^2}{h^2/\ell^2}=\frac{2gh\ell^2+2gh^3}{h^2}=2g\ell^2/h+2gh,$$
Therefore $$v_o=\sqrt{2gh+2g\ell^2/h}.$$
I would look at the solution, but the solution does not solve the problem "from scratch", instead it looks at special cases of choosing $h \rightarrow 0$ and $\ell \rightarrow 0$ limits. Where did I go wrong in my proof? I've checked my initial projectile motion equations and they're correct.
Best Answer
Your approach is incorrect from the point that you say that $\theta = \arctan(h/\ell)$. Here's a derivation you might like, though.
The $y$-component of the velocity must satisfy $h = \frac{v_y^2}{2g} \implies v_y = \sqrt{2gh}$. We must now choose an $x$-component $v_x$ such that the total horizontal distance traveled is $\ell$. The time required for the ball to reach the top of its arc will be given by $h = \frac 12gt^2 \implies t = \sqrt{2h/g}$. Thus, $v_x$ must satisfy $v_x = \ell/t = \ell \sqrt{\frac{g}{2h}}$.
Now, calculate the speed as $\|v\| = \sqrt{v_x^2 + v_y^2}$.