Can You Maximize The Turkey Trajectory?

Riddler Classic

Trig the turkey cannot fly. And so, instead of flying, he decides to jump as far as he can with a running start up one of the famed Sinusoidal Hills, all of which are precisely the same size.

Trig knows that he cannot jump a horizontal distance of more than two hills. Also, he prefers a smooth takeoff and landing. That is, when he takes off from the ground and lands on it again, the slope of his parabolic trajectory through the air must perfectly match the instantaneous slope of the ground beneath him.

The animation below shows a jump where the takeoff and landing are precisely 1.2 hills apart.

What is the greatest horizontal distance Trig can jump, such that his takeoff and landing are smooth? Again, keep in mind that Trig cannot possibly jump a horizontal distance greater than two hills.

Solution

Let $H(x) = \cos {\left(\pi x\right)}$ represent the hills. Two peaks are at $(-1, 1)$ and $(1, 1)$, and the trough in between is at $(0, -1)$.

Centering Trig's parabolic trajectory with the hills, let his jump be represented by $T(x) = ax^2+c$

Since the instantenous slopes have to match,
$$$H'(x) = T'(x)$$
$$-\pi\sin {\left(\pi x\right)} = 2ax$$
$$a = \dfrac{-\pi\sin {\left(\pi x\right)}}{2x}$$
In order to prevent Trig's parabolic trajectory to not intersect the upward slope of the hill, take the derivative of the above and set to 0.
$$0 = \dfrac{2x\left(-\pi^2\cos {\left(\pi x\right)}\right)-\big(-\pi\sin {\left(\pi x\right)}\big)(2)}{4x^2}$$
$$0 = -x\big(\pi\cos {\left(\pi x\right)}\big)+\sin {\left(\pi x\right)}$$
$$x\pi\cos {\left(\pi x\right)}=\sin {\left(\pi x\right)}$$
$$x\pi=\tan {\left(\pi x\right)}$$

Answer

Wolfram Alpha gives the solution as $x = 1.43029665312420...$

Rohan Lewis

2021.11.29