#### Answer

(a) $a = \frac{gh}{d}$
(b) $a = 70~m/s^2$

#### Work Step by Step

(a) We can find the required speed at the top of the tube as:
$v^2 = 0+2gh$
$v = \sqrt{2gh}$
We can find the required acceleration in the tube as:
$a = \frac{v^2-v_0^2}{2d}$
$a = \frac{(\sqrt{2gh})^2-0}{2d}$
$a = \frac{gh}{d}$
(b) $a = \frac{gh}{d}$
$a = \frac{(9.80~m/s^2)(3.2~m)}{0.45~m}$
$a = 70~m/s^2$