$解：?(1) ?由題意，可知拋物線頂點(diǎn)?D?的坐標(biāo)為?(12，??20)，?點(diǎn)?B?的坐標(biāo)為?(0，??2)?$
$∴設(shè)拋物線相應(yīng)的函數(shù)表達(dá)式為?y=a(x-h)^2+k，?即?y=a(x-12)^2+20?$
$∵點(diǎn)?B?在拋物線上$
$∴?2=a(0-12)^2+20，?即?a=- \frac {1}{8}?$
$∴該拋物線相應(yīng)的函數(shù)表達(dá)式為：?y=- \frac {1}{8} x^2+3x+2(0≤x≤12+4 \sqrt{10} ) ?$
$?(2)?過點(diǎn)?C?作?CE⊥x?軸，垂足為?E?$
$設(shè)?CE=b，??AE=a?$
$則? \begin{cases}{tanβ =\dfrac {a}=\dfrac {2}{3}}\\{tanα=\dfrac b{a+2}=\dfrac 35}\end{cases}，?解得?\begin{cases}{a=18}\\{b=12}\end{cases}?$
$則點(diǎn)?C?的坐標(biāo)為?(20，??12)?$
$當(dāng)?x=20?時，函數(shù)值?y=- \frac {1}{8} ×20^2+3×20+2=12?$
$∴能點(diǎn)燃目標(biāo)?C?$
