Answer
$108\;\rm m$
Work Step by Step
First, we need to find his velocity components $v_{x'}$ just before he leaves the slope; and to do so we need to draw the force diagram on him, as shown in the figures below.
From the first figure above;
$$\sum F_{x'}=F_T-mg\sin\theta=ma_{x'}$$
Thus,
$$F_T-mg\sin\theta=ma_{x'}$$
Solving for $a_{x'}$;
$$ a_{x'}=\dfrac{F_T-mg\sin\theta}{m}=\dfrac{200-(75\times 9.8\sin10^\circ)}{75}$$
$$a_{x'}=\color{blue}{\bf 0.965}\;\rm m/s^2$$
Now we need to find $v_{x'}$;
$$v_{x'}^2=\overbrace{v_{0x'}^2}^{0}+2a_{x'}d$$
$$v_{x'} =\sqrt{\overbrace{v_{0x'}^2}^{0}+2a_{x'}d}$$
$$v_{x'} =\sqrt{ 2a_{x'}d}$$
where $d$ is given by $d=\dfrac{h}{\tan\theta}$, as shown in the second figure below.
$$v_{x'} =\sqrt{ 2a_{x'}\dfrac{h}{\tan\theta} }$$
Plugging the known;
$$v_{x'} =\sqrt{ 2\times 0.965\times \dfrac{50}{\tan10^\circ} }=\color{blue}{\bf 23.4}\;\rm m/s $$
Now we need to find the two velocity components of $v_{x'}$ due to the new coordinate which are the normal coordinates of $x$ and $y$.
$$v_{x}=v_{x'} \cos10^\circ =23.4\cos10^\circ =\color{magenta}{\bf 23.9}\;\rm m/s $$
$$v_{y}=v_{x'} \sin 10^\circ =23.4\sin 10^\circ =\color{magenta}{\bf 4.1}\;\rm m/s $$
Now we need to find the time of his trip until the moment he touches the ground.
And to do that we need to find his vertical acceleration component (we also need to find the horizontal acceleration component).
From the last figure below;
$$\sum F_y=F_T\sin\theta-mg=ma_y$$
Thus,
$$a_y=\dfrac{F_T\sin\theta-mg}{m}=\dfrac{200\sin10^\circ-(75\times 9.8)}{75}$$
$$a_y=\bf-9.34\;\rm m/s^2$$
$$\sum F_x=F_T\cos\theta =ma_x$$
Thus,
$$a_x=\dfrac{F_T\cos\theta }{m}=\dfrac{200\cos10^\circ }{75}$$
$$a_x=\bf2.63\;\rm m/s^2$$
Now to find how far he lands from the base of the slope, we need to find the time of his vertical trip.
$$\overbrace{y}^{=0}-\overbrace{y_1}^{=h}=v_{y}t+\frac{1}{2}a_yt^2$$
Plugging the known and solving for $t$;
$$-50=4.1t+\frac{-9.34}{2} t^2$$
$$4.67t^2-4.1t-50=0$$
So, $t=-2.86$ s; which is dismissed or $t=\bf 3.74$ s.
Thus, the horizontal distance traveled is given by
$$x=\overbrace{x_0}^{=0}+v_xt+\frac{1}{2}a_xt^2$$
Plugging the known;
$$x= (23.9\times 3.74)+\left(\frac{1}{2}\times 2.63\times 3.74^2\right)$$
$$x=107.8\;\rm m\approx \color{red}{\bf 108}\;m$$
