Short Ride in a Fast Machine

The mass, angle, speed, gravitational acceleration and types of dissipated energy are all irrelevant.

Since $E = \frac{1}{2}mv^{2}$ , for the total energy to be $\frac{2}{3}$ of the original while keeping mass constant, $v$ has to change. At the first landing, the speed becomes $\sqrt{\frac{2}{3}}u$ . Let's call this factor $\gamma = \sqrt{\frac{2}{3}}$ .

The angle still remains $\theta$ (see the comments on why).

At the $n$ th landing, the speed is $\gamma^{n}u$ . Since the range of a projectile is known to be $\displaystyle \frac{u^{2}\sin 2\theta}{g}$ . Considering all the ranges, we obtain the sum

$\sum_{n=0}^{\infty} \frac{\gamma^{2n}u^{2}\sin 2\theta}{g} = \frac{u^{2}\sin 2\theta}{g} \frac{1}{1-\gamma^{2}} = \frac{3u^{2}\sin 2\theta}{g}$

The ratio of this to the first range, $\displaystyle \frac{u^{2}\sin 2\theta}{g}$ , is clearly, then, $\boxed{3}$ .