Focal length of the mirror shown in figure is 20$cm$. Find the image position and its velocity.

Substituting the values,

$u=-30cm and f=-20cm$

in the mirror formula, we have 

$\frac{1}{v}+\frac{1}{-30}=\frac{1}{-20}$

solving this equation, we get $v=-60cm$.

Further, $$\begin{gathered} m=-\frac{v}{u}=-\frac{\left(-60\right)}{\left(-30\right)}=-2 \\ and m^2=4 \end{gathered}$$

Object velocity along the axis is $5\cos 37° = 4 mm/s$ (towards $OP$). Therefore, image velocity along the axis should be $m^2$ times or $16 mm/s$ in the opposite direction of object velocity. Object velocity perpendicular to axis is $5\sin 37° = 3 mm/s$ (upwards). Therefore, image velocity will be $m$ times or $-6 mm/s$ downwards. The position and velocity of image is shown below.

$$\begin{gathered} v_I=\sqrt{{\left(16\right)}^2+{\left(6\right)}^2}=\sqrt{292} mm/s \\ \tan \theta =\frac{6}{16}=\frac{3}{8} or \theta ={\tan }^{-1}\left(\frac{3}{8}\right) \end{gathered}$$