$解：?(1)?把?x=0，??y= 2?及?h= 2.6?代入?y= a(x-6)2+h?$
$即?2=a(0-6)2+2.6?$
$∴?a=-\frac {1}{60}?$
$當(dāng)?h= 2.6?時，?y?與?x?的函數(shù)表達式為?y=-\frac {1}{60}(x-6)2+ 2.6 ?$
$?(2)?當(dāng)?h= 2.6?時，?y=-\frac {1}{60}(x- 6)2+2.6?$
$∴當(dāng)?x=9?時；$
$?y=-\frac {1}{60}(9- 6)2+ 2.6= 2.45\gt 2.43?$
∴球能越過網(wǎng)
$∵當(dāng)?y= 0?時，即?-\frac {1}{60}(x-6)2+2.6= 0?$
$解得?{x}_1=6+\sqrt{156}\gt 18，??{x}_2= 6-\sqrt{156}(?不合題意，舍去)$
$或當(dāng)?x = 18?時，?y=-\frac {1}{60}(18- 6)2+ 2.6= 0.2\gt 0?$
∴球會出界
$?(3)?把?x=0，??y= 2，?代入?y=a(x-6)2+h ?得?a=\frac {2-h}{36}?$
$當(dāng)?x=9?時，?y=\frac {2-h}{36}×(9-6)2+h=\frac {2+3h}{4}＞2.43①?$
$當(dāng)?x=18?時，?y=\frac {2-h}{36}×(18-6)2+h=8-3h≤0 ②?$
$由①②解得?h≥\frac {8}{3}。?$
$∴若球一 定能越過球網(wǎng)，又不出邊界，?h ?的取值范圍為?h≥\frac {8}{3}?$
