$$
s=\frac{b}{2} t^{2}, \quad s=\frac{c}{\cos \alpha}, \quad \text { also } \quad \frac{c}{\cos \alpha}=\frac{b}{2} t^{2} \quad \text { or } \quad t=\sqrt{\frac{2 c}{b \cos \alpha}}
$$

From the last equation, the acceleration $b$ still needs to be eliminated. From $F=m b$ it follows that $b=\frac{F}{m}$ and since $F=G-\sin \alpha$ and $G=g \cdot m$, we get $b=g \cdot \sin \alpha$. Therefore, using an addition theorem,

$$
t=\sqrt{\frac{2 c}{g \cdot \sin \alpha \cos \alpha}}=\sqrt{\frac{4 c}{g \cdot \sin (2 \alpha)}}
$$

The time $t$ should assume a minimum value; this is exactly the case when the denominator of the fraction under the square root assumes a maximum value, i.e., when $\sin (2 \alpha)=1$. This immediately yields

$$
\alpha=45^{\circ} ; \quad t_{\min }=\sqrt{\frac{4 c}{g}}=2 \sqrt{\frac{c}{g}}
$$
## Task 35/62

What inclination angle $\alpha$ must a sloping plane with base $c$ have so that a sphere rolling down it does so in the shortest time? Friction and the moment of inertia are neglected.