Recent content by avenior

##\operatorname{tg}\varphi^{'} = \frac{u \sin\varphi}{\gamma (u \cos\varphi - V)}## and ##\varphi = \frac{\pi}{2}##. Hence tangent of the boundary angle in a moving coordinate system ##\operatorname{tg}\varphi^{'} = -\frac{1}{\gamma}\frac{u}{V}##. But ##\frac{\alpha^{'}}{2} = \pi - \varphi^{'}##...

Yes. Given the condition of the task, I think ##\varphi \in \left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]##.

##\varphi \in \left[0,\pi\right]## or ##\varphi \in \left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]##?

##\cos \varphi < \frac{V}{u}## and ##\varphi## should depend on ##\frac{\alpha}{2}##, but I don't understand how to take into account half of the splinters.

Ok, thanks. But how does it relate to the fact that half should collide with the surface?

The splinter will collide with the moving surface if $$-\operatorname{tg}\frac{\alpha}{2} < \frac{u \sin\varphi}{\gamma (u \cos\varphi - V)} < \operatorname{tg}\frac{\alpha}{2}$$

Homework Statement On the axis of an infinite wedge that moves with velocity ##\vec{V}##, the body decays with the formation of a lot of splinters that fly away uniformly in all directions with velocity ##\vec{u}##. What should be the angle of the wedge that half of the splinters fall on its...