= Solution
The <arc length> of the <helix> follows from its parametrization:
$$
\mathbf r'(t)=\bigl(-10\sin(t-\pi/2),\ 10\cos(t-\pi/2),\ 5\bigr),\qquad
|\mathbf r'(t)|=\sqrt{100+25}=5\sqrt5.
$$
Thus $\ell_h=15\pi\sqrt5\,\mathrm m$. Its endpoint is $(0,20,15\pi)$, at height $h=15\pi\,\mathrm m$.
Under the usual intended interpretation that the descending quarter of a <circle> drops through one <radius>, its <radius> is $R=h$. Its <arc length> is $\ell_c=\pi R/2=15\pi^2/2\,\mathrm m$, giving
$$
\boxed{\ell_h+\ell_c=15\pi\sqrt5+\frac{15\pi^2}{2}=179.39425397\ldots\,\mathrm m\simeq179\,\mathrm m.}
$$
\b[The printed smoothness condition is inconsistent.] The terminal <tangent vector> of the <helix> is $\mathbf r'(3\pi)=(-10,0,5)$. Every <tangent vector> to a <curve> confined to the $yz$ <plane> has zero $x$ component, so no such <regular curve> can join this <helix> with a common <tangent line>. Moreover, the <helix> is still rising at that endpoint. Thus no smooth ride satisfying all the literal conditions exists.
Even after allowing a corner, a quarter of a <circle> reaching the ground does not alone determine its <radius>: the orientation of its endpoint radii must also be specified. For example, in the $yz$ <plane>, the circular arc
$$
y=20+R\bigl(\cos\theta-\cos\theta_0\bigr),\quad
z=h+R\bigl(\sin\theta-\sin\theta_0\bigr),\quad
\theta_0-\pi/2\le\theta\le\theta_0,
$$
traversed toward decreasing $\theta$, descends monotonically for $0\le\theta_0\le\pi/2$ and reaches $z=0$ when $R=h/(\sin\theta_0+\cos\theta_0)$. Choosing $\theta_0=\pi/4$ gives $R=h/\sqrt2$, whereas $\theta_0=\pi/2$ gives $R=h$. The numerical answer above is conditional on the intended $R=h$ geometry; it does not repair the impossible smooth join.
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