A round cone with half-angle  $\alpha ={45}^0$ and the radius of the base R = 10 cm rolls uniformly without slipping over horizontal plane. The apex is hinged at the point O (see figure) which is at the same level as the point C, the centre of the base. The velocity of the point C is  $v=10m{s}^{-1}$

There are two angular motions of the cone, one in the horizontal plane with angular velocity ${\omega }_1=\frac{v}{R\cot \alpha }$  where $R\cot \alpha =$  radius of the circle in which c moves. The direction of this vector is upward. The other angular velocity ${\omega }_2$  is about its own axis. Since it rolls without slipping,$v={\omega }_{2}R\Rightarrow {\omega }_2=\frac{v}{R}$  . The direction of this vector is horizontal and towards O.
   ${\omega }_{results}=\sqrt{{\omega }_1^2+{\omega }_2^2}=\sqrt{\frac{v^2}{R}+\frac{v^2}{R}{\tan }^2\alpha ;}$$\frac{v}{R}\sqrt{1+{\tan }^2\alpha }=\frac{v}{R}\sec \alpha$
Substituting the values of  $V\&R100\sqrt{2}$
Let $\theta$  be the angle made by the resultant with the vertical. Then $\tan \theta =\frac{{\omega }_2}{{\omega }_1}\theta ={45}^0$