The three constraints outlined in the Understanding the Problem section above define this problem and hence its solution. We will put them to good use:

The first thing to realize is that for any jump from \(x_0\) that satisfies the first condition outlined in the Understanding the question section above, Trig has to land on Hill 2 a distance \(x_0\) from the end of Hill i.e. from \(4\pi \), giving us a range R

\[
\color{Red}{R = 4\pi – 2x_0  \tag{Range} \label{Range}}
\]

This is not handwaving but a consequence of the symmetry of projectile motion and the argument that were this not so, we could always increase Trig’s jump range by making the jump symmetric around the hills. (just think about it). This range obviously increases as \(x_0\) decreases. So can we just not lift off at \(x_0 = 0\)? Well, obviously not exactly 0 since Trig doesn’t get anywhere without some lift. So how much higher up the hill should he go before liftoff? 

The second thing Trig’s jump has to satisfy is that the slopes of the Hill curve and Trig’s curve have to be identical at the (symmetrical) points of liftoff and landing. Derivatives give us slopes. Now the slopes of H and T above are:

\begin{align*}
\frac{dH}{dx} &= \sin x \\
\frac{dT}{dx} &= a – 2bx \\
\end{align*}

Equating these at \(x_0\) and \(4\pi – x_0\) gives us:

\begin{align*}
\sin x_0 &= a – 2bx_0 \\
\sin (4\pi – x_0) &= a -2b(4\pi -x_0) \qquad \\ 
\end{align*}

which can be used to eliminate \(a\) to get 

\[
-2b = \frac{\sin x_0}{x_0 – 2\pi }
\]

You might note now that neither the Range equation nor the result above prevent us from launching at say just over \(x_0\) for a range of just under 2 Hill length or \(45^{\large\circ} \) for a range of 1.5 Hill lengths. What does restrict us is the need for Trig to clear the first Hill without slamming into it at any point. Which brings us to…

The third and final thing to complete our solution is that for Trig to clear Hill 1, the hill ground below Trig has to fall off faster than his parabolic trajectory. This means the rate at which the slope of the hill falls has to be greater than the rate at which Trig’s trajectory’s slope drops. Second derivatives! Now the rates of change of the slopes are:

\begin{align*}
\frac{d^2 H}{dx^2} &= \cos x \\
\frac{d^2 T}{dx^2} &= -2b \qquad \textit{a negative constant!} \\
\end{align*}

and clearing the hill at a chosen \(x_0\) implies:

\begin{align*}
-2b &\geq cos x_0 \\
\frac{\sin x_0}{x_0 – 2\pi} &\geq \cos x_0 \qquad \textit{using value of } -2b \textit{ from above} \\
\Rightarrow \qquad \tan x_0 &\geq x_0 – 2\pi \qquad \textit{(}x_0 < \frac{\pi}{2} \textit{ so sin and cos are positive)} \\
\Rightarrow \qquad x_0 &\geq 1.78978 \qquad \textit{solved numerically}
\end{align*} 

Plugging this value of \(x_0\) into our equation \(\eqref{Range} \) above, we get:

\[
R = 4\pi -2(1.78978) = 8.99
\]

The final answer in terms of Hill units: since 1 hill length = \(2\pi \) 

\[ \bbox[yellow, 10px]{\displaystyle{\frac{8.99}{2\pi}} = 1.43 \textit{ hill units}} \]

We can see that this is greater than 0 and larger than  \(\displaystyle{\frac{\pi}{4}}\) or beyond the \(45^{\large\circ} \) launch point textbook answer for maximum projectile range. Liftoff any earlier than 1.79 and  Trig slams into Hill1 at some point and instantly becomes a Thanksgiving meal centerpiece (Trig cannot jump more than 2 Hills so his speed is restricted). 

So 1.43 hill units is the best Trig can do under the circumstances. Trig’s flight is shown in the figure below. The liftoff angle is a tad lower than \(45^{\large\circ} \) at \(44.3^{\large\circ} \) as expected. 

Note that this answer is independent of  the actual width of the hills and any specific value of gravity etc. So this result would even hold if Trig were to ever try this stunt on the Moon, or Mars (in an appropriate spacesuit of course)!

Edit: Someone asked what if the hill height to width ratio were different. I haven’t shown this, but the answer does not change. To convince yourself just plug in a factor \(k\) into the hill Cosine equation and work through all the same steps to see that the \(k\) cancels out!