= Solution
Neglect condensation and the smaller droplets' fall speed, while retaining their specified liquid-water content. For <exponential cloud collector growth>, the collection law and spherical collector <mass> give
$$
4\pi\rho_lR^2\dot R=\pi R^2(k_RR)LE,\quad
\dot R=\alpha_RR,\qquad\alpha_R=\frac{k_RLE}{4\rho_l}=1.6\times10^{-3}\,\mathrm{s^{-1}}.
$$
Thus $R(t)=R_0e^{\alpha_Rt}$. The growing terminal speed eventually exceeds the updraft, and
$$
\boxed{h_R(t)=Ut-\frac{k_RR_0}{\alpha_R}(e^{\alpha_Rt}-1)
=t\,\mathrm{m\,s^{-1}}-100\,\mathrm m\,(e^{0.0016t}-1),}
$$
where the last expression uses $t$ in seconds and $U=1\,\mathrm{m\,s^{-1}}$. Numerically,
$$
\boxed{h_R(1000\,\mathrm s)=605\,\mathrm m,\qquad h_R(2000\,\mathrm s)=-353\,\mathrm m\quad\text{(formal extrapolation)}.}
$$
The trajectory reaches maximum height when $k_RR=U$, at $t=1145\,\mathrm s$, $R=125\,\mu\mathrm m$ and $h_{\max}=620\,\mathrm m$. It returns to cloud base near $t=1860\,\mathrm s$, with radius about $392\,\mu\mathrm m$. The left-to-right sketch in the preceding figure shows this rise and subsequent fall.
Once the drop leaves the cloud, the prescribed in-cloud $L$, $U$ and collection law need not persist. Consequently the negative $2000$-second value describes the requested constant-coefficient model's extension, not a prediction that collection continues below cloud base. Its actual subsequent height needs an outside-cloud model. Within the cloud, the exponential feedback between increasing radius and increasing swept volume explains how a minority of collector drops can escape the condensation bottleneck and initiate warm rain; depletion, breakup and changes in fall law eventually limit indefinite exponential growth.
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