Answer
No.
Work Step by Step
We need to divide his motion into two stages, the first stage is when he is moving up the incline, and the second stage is when he is in the air under the free-fall acceleration (air resistance is neglected).
Each stage has its own coordinates, as we see in the figures below. So do not be confused about these two coordinates we do that to make the solution easier.
To find out if he survives, or not, we need to find his speed at the end of the ramp then we can find the distance traveled by him in the horizontal direction after that during his time of falling.
We need to find his acceleration while he is moving up the incline, so we need to draw the force diagram of him and we also need to analyze his weight to its $x$ and $y$ components, as shown below.
The distance traveled by the motorcyclist up the incline $d$ is given by
$$ \sin\theta=\dfrac{h}{d}$$
thus,
$$d=\dfrac{h}{\sin\theta} \tag 1$$
And hence,
$$v_{x1}^2=v_{x0}^2+2a_xd$$
$$v_{x1}^2=(v_{0}\cos\theta)^2+2a_xd$$
Plugging from (1);
$$v_{x1}^2=(v_{0}\cos\theta)^2+2a_x\dfrac{h}{\sin\theta}\tag 2$$
From the second figure below,
$$\sum F_x= -f_r-mg\sin\theta=ma_x$$
where the rolling friction is given by $f_r=\mu_rF_n$
$$-\mu_rF_n-mg\sin\theta=ma_x\tag 3$$
And
$$\sum F_y= f_n-mg\cos\theta=ma_y=m(0)=0$$
Thus,
$$ f_n=mg\cos\theta$$
Plugging into (3);
$$-\mu_r \color{red}{\bf\not}mg\cos\theta- \color{red}{\bf\not}mg\sin\theta= \color{red}{\bf\not}ma_x $$
$$a_x=-\mu_r g\cos\theta- g\sin\theta=-g(\mu_r \cos\theta+\sin\theta)$$
Plugging into (2);
$$v_{x1}^2=(v_{0}\cos\theta)^2-2g(\mu_r \cos\theta+\sin\theta) \dfrac{h}{\sin\theta} $$
$$v_{x1} =\sqrt{(v_{0}\cos\theta)^2-(\mu_r \cos\theta+\sin\theta) \dfrac{2gh}{\sin\theta}} $$
Plugging the known;
$$v_{x1} =\sqrt{(11\cos20^\circ)^2-(0.02 \cos20^\circ+\sin20^\circ) \dfrac{2\times 9.8\times 2}{\sin20^\circ}} $$
$$v_{x1}=\color{blue}{\bf 8.093}\;\rm m/s$$
Now we can find his vertical velocity component $v_{iy'}$ at the first moment of his second stage of motion;
$$v_{iy'}=v_{x1}\sin\theta= 8.093\sin20^\circ=\color{green}{\bf 2.77}\;\rm m/s$$
The time he takes to reach the ground is given by
$$\overbrace{y'_f}^{=0}-\overbrace{y'_0}^{=2}=v_{iy'}t+\frac{1}{2}\overbrace{a_{y'}}^{=-g}t^2$$
Plugging the known and solving for $t$;
$$-2=2.77t-\frac{1}{2}gt^2$$
$$4.9t^2-2.77t-2=0$$
Thus, $t=-0.416\;\rm s$ which is neglected, or $t=0.981\;\rm s$.
Now we need to find the distance traveled in the horizontal direction during this time of 0.981 s.
So, first, we need to find the horizontal velocity component;
$$v_{ix'}=v_{x1}\cos\theta= 8.093\cos20^\circ=\color{green}{\bf 7.6}\;\rm m/s$$
The horizontal distance is given by
$$\Delta x'=v_{ix'}t$$
since his horizontal acceleration component is zero.
$$\Delta x'=7.6\times 0.981=\color{red}{\bf 7.5}\;\rm m$$
Therefore, it seems that the crocodiles will have fresh meat for dinner.
