/sci/ - Science & Math

>>10124488
It's basically like >>10123052 said. However, you have to take into account that the ball will lose contact with the boxes. We get
[eqn] \frac{1}{4}mv^2 = -\frac{mgy}{2} = \frac{m}{2}g(1-\cos \varphi)\frac{D}{2} \\
v^2 = gD(1-\cos \varphi) [/eqn]
Normal force
[eqn] F_N = -\frac{2mv^2}{D} + mg\cos \varphi = mg(3\cos \varphi - 2) [/eqn]
The ball and boxes lose contact when F_N is 0 and thus [math]\varphi_0 = \arccos{\frac{2}{3}}[/math]. The velocity of the point in the vertical direction is [math]v_y(\varphi) = v(\varphi)\sin \varphi[/math]

Torricelli's equation
[math] v^2_{y,\mathrm{End}} = v^2_y(\varphi_0) + 2g\Delta y [/math], with [math] \Delta y = D-\frac{D}{2}(1-\cos \varphi_0) [/math]

Plug everything in
[math] v^2_{y,\mathrm{End}} = gD(1-\cos(\arccos\frac{2}{3}))(1-(\frac{2}{3})^2) + 2g(\frac{5}{6}D) = \frac{5}{27}gD + \frac{10}{6}gD = \frac{50}{27}gD [/math]