Frog is jumping infinitively - hard

Let each parabola $y_n$ (starting from $n=0$ ) be centred at the origin with the vertex at $(0,h_n)$ where $h_n$ is the maximum height of the jump, and cutting the x-axis at $\frac {l_n}2$ and $-\frac {l_n}2$ where $l_n$ is the horizontal length of the jump. This gives each parabola the following equation, which can be found by letting $y=ax^2+b$ and finding appropriate values for $a$ and $b$ such that the above conditions hold. $y_n=h_n\left(1-\frac{4x^2}{l_n^2}\right)$ $l_0=4$ , and $l_n=\frac34l_{n-1}$ , therefore $l_n=4\left(\frac34\right)^n$

$h_0=1$ , and $h_n=\frac23h_{n-1}$ , therefore $h_n=\left(\frac23\right)^n$ $\begin{aligned}\therefore y_n&=\left(\frac23\right)^n\left(1-\frac{4x^2}{\left(4\left(\frac34\right)^n\right)^2}\right)\\ &=\left(\frac23\right)^n-\left(\frac23\right)^n\left(\frac{16}9\right)^n\frac{4x^2}{16}\\ &=\left(\frac23\right)^n-\left(\frac{32}{27}\right)^n\frac{x^2}{4}\end{aligned}$

Now, the formula for arc length is $\int_a^b\sqrt{(f'(x))^2+1}dx$ . Call the length of the trajectory of one parabola $T_n$ . $\begin{aligned}T_n&=\int_{-\frac{l_n}2}^{\frac{l_n}2}\sqrt{\left(\frac{dy_n}{dx}\right)^2+1}dx\\ &=2\int_0^{\frac{l_n}2}\sqrt{\left(-\left(\frac{32}{27}\right)^n\frac{x}{2}\right)^2+1}dx\\ &=2\int_0^{2\left(\frac34\right)^n}\sqrt{\left(\frac{32}{27}\right)^{2n}\frac{x^2}{4}+1}\\ &=\left(\frac{32}{27}\right)^n\int_0^{2\left(\frac34\right)^n}\sqrt{x^2+4\left(\frac{27}{32}\right)^{2n}}\\ &=\left(\frac{32}{27}\right)^n\left[\frac x2\sqrt{x^2+4\left(\frac{27}{32}\right)^{2n}}+2\left(\frac{27}{32}\right)^{2n}\ln\left|x+\sqrt{x^2+4\left(\frac{27}{32}\right)^{2n}}\right|\right]_0^{2\left(\frac34\right)^n}\quad\small\text{using }\int\sqrt{x^2+a^2}=\frac x2\sqrt{x^2+a^2}+\frac{a^2}2\ln\left|x+\sqrt{x^2+a^2}\right|\\ &=\left(\frac34\right)^n\left(\frac{32}{27}\right)^n\sqrt{4\left(\frac9{16}\right)^n+4\left(\frac{27}{32}\right)^{2n}}+2\left(\frac{27}{32}\right)^n\ln\left(2\left(\frac34\right)^n+\sqrt{4\left(\frac9{16}\right)^n+4\left(\frac{27}{32}\right)^{2n}}\right)-2\left(\frac{27}{32}\right)^n\ln\left(2\left(\frac{27}{32}\right)^n\right)\\ &=2\left(\frac34\right)^n\sqrt{1+\left(\frac98\right)^{2n}}+2\left(\frac{27}{32}\right)^n\ln\left(\left(\frac89\right)^n+\sqrt{1+\left(\frac89\right)^{2n}}\right)\end{aligned}$

The total length of the trajectory of the frog will be $\displaystyle\sum_{n=0}^\infty T_n$ , but since this is rather complicated we may just evaluate this with a computer program, and get a result of $\boxed{17.489}$