If you look at the problem from a realistic point of view, it depends on the conditions of system. With energy losses from air resistance and the energy lost during the rebound, the maximum height of the ball will decrease over time, which makes it much more likely that the bear character will not reach the other side. However, note that the bear can reach the other side in many different ways. If it is strong enough, it could traverse the canyon with a single jump, for example.
But the picture has some implicit assumptions that one can use to solve the problem quantitatively:
* The bear goes to the other side by using many jumps along the way
* The height at which each jump occurs is the same
* The movement is regular and periodic, which suggests that energy losses are not taken into account.
With that in mind, it is possible to solve for the momentum the bear has to give to the ball.
Because the ball and the bear move together horizontally, the bear cannot give the ball an impulse horizontally. If it propels itself forward, the ball will be thrown in the opposite direction, and the bear will not be able to jump on it anymore. However, that is not an issue with no energy losses. 621
If the bear and the ball start with an initial horizontal velocity, that will be maintained as long as the bear only exert force on the ball vertically. In this sense, we only need to consider the vertical motion.
At the very beginning, both the ball and the bear are mid-air with a constant horizontal velocity. To not fall, the bear jumps such that it gains a momentum $Q$ and the ball a momentum $-Q$, due to conservation of momentum. Then, the magnitude of the velocity of the bear and the ball are, respectively, $$V=\frac{Q}{M}$$ $$v=\frac{Q}{m}$$ Then, the time $t$ the bear takes to return to the same height (the height immediately after the moment it jumped) is given by the following equation of motion $$Vt-\frac{1}{2}gt^2+h=h$$ $$t=\frac{2V}{g}$$ The time $t_1$ the ball takes to reach the ground is given by $$-vt_1-\frac{1}{2}gt_1^2+h=0$$ $$t_1=\frac{-v+\sqrt{v^2+2gh}}{g}$$ where $h$ is the height of the canyon/valley. The ball reaches the ground with a velocity $$v_f=\sqrt{v^2+2gh}$$ The time $t_2$ the ball takes to return to the same height is given by $$v_ft_2-\frac{1}{2}gt_2^2=h$$ $$t_2=\frac{v_f+\sqrt{v_f^2-2gh}}{g}=\frac{\sqrt{v^2+2gh}-v}{g}$$ By using the fact that when the jump occurs the bear and ball meet, then $$t=t_1+t_2$$ $$V=\sqrt{v^2+2gh}-v$$ $$\frac{Q}{M}=\sqrt{\frac{Q^2}{m^2}+2gh}-\frac{Q}{m}$$ Solving for $Q$: $$Q=\frac{M\sqrt{2mgh}}{\sqrt{2M+m}}$$ Which has a real solution. Now, this was for the first jump, but this can be repeated as many times as necessary, because the total momentum immediately before the second jump is 0, just like immediately before the first jump, because the ball and the bear come back to the same height with the same velocities, just with opposites directions.
Using $M=200kg$, $m=1kg$, $g=9.8m/s²$, $h=10m$, then $$v\approx 140 m/s$$ 771