## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$a = \frac{gh}{d}$
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: $a = \frac{v^2-v_0^2}{2d}$ $a = \frac{(\sqrt{2gh})^2-0}{2d}$ $a = \frac{gh}{d}$