Chapter 2 - Kinematics in One Dimension - Exercises and Problems - Page 66: 61

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

(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) $a = \frac{gh}{d}$ $a = \frac{(9.80~m/s^2)(3.2~m)}{0.45~m}$ $a = 70~m/s^2$