Let the point where the water slide touches the ground be $D$. At point $D$, $v_0^2=2gH$.
At point $B$, $v^2=v_0^2-2gh=2g(H-h)$.
Now, Azmain should travel on
PROJECTILE MOTION.(If you have no idea about projectile motion, you can click on the
link for easy explanation.)
Velocity among $X$ axis, $v_x=vcosθ$.
Velocity among $Y$ axis, $v_y=vsinθ$.
Now, total time required, $t=t_1+t_2=\dfrac{2vsinθ}{g}+\dfrac{\sqrt{v^2sin^2θ+2gh}-vsinθ}{g}=\dfrac{vsinθ+\sqrt{v^2sin^2θ+2gh}}{g}$
So, distance from point $C$,
$d=v_xt=vcosθ\dfrac{vsinθ+\sqrt{v^2sin^2θ+2gh}}{g}=\sqrt{2g(H-h)}cosθ\dfrac{\sqrt{2g(H-h)}sinθ+\sqrt{2g(H-h)sin^2θ+2gh}}{g}=\dfrac{2g(H-h)cosθ sinθ+\sqrt{4g^2(H-h)^2sin^2θ cos^2θ+4g^2(H-h)h cos^2θ}}{g}=\dfrac{2g(H-h)cosθ sinθ+2g\sqrt{(H-h)^2sin^2θ cos^2θ+(H-h)h cos^2θ}}{g}=2\left((H-h)cosθ sinθ+\sqrt{(H-h)^2sin^2θ cos^2θ+(H-h)h cos^2θ}\right)=2cosθ\left((H-h) sinθ+\sqrt{(H-h)^2sin^2θ +(H-h)h}\right)=2cosθ\left((H-h) sinθ+\sqrt{(H-h)(H sin^2θ-hsin^2θ +h)}\right)=2cosθ\left((H-h) sinθ+\sqrt{(H-h)(H sin^2θ+h(1-sin^2θ))}\right)=2cosθ\left((H-h) sinθ+\sqrt{(H-h)(H sin^2θ+h cos^2θ)}\right)$
The interesting fact is that distance of the system doesn't depend on gravity. That means, Azmain would travel same distance in both the Earth and the moon, if there was no wind resistance.