$解:(1)根據(jù)題意可得:拋物線過點(diǎn)A(0,10),(3,7),對(duì)稱軸為x=1$
$所以設(shè)拋物線的解析式為y=ax2+bx+10$
$\begin{cases}{ 9a+3b+10=7 } \\ {-\frac{2a}=7} \end{cases}$
$解得a=-1,b=2$
$所以y關(guān)于x的函數(shù)解析式為y=-x2+2x+10$
$(2)令y=0,則0=-x2+2x+10$
$解得x_1=\sqrt{11}+1,x_2=-\sqrt{11}+1(不合題意,舍去)$
$所以O(shè)B的長為\sqrt{11}+1m.$
