A ball is kept at a height h above the surface of a heavy transparent sphere made of a material of refractive index $\mu$ . The radius of the sphere is R. At t = 0, the ball is dropped to fall normally on the sphere. For $\mathrm{t}<\sqrt{\frac{2\mathrm{h}}{\mathrm{g}}}$ and by a single refraction, what is the speed of image as a function of time ? 

$$\begin{gathered} \mathrm{AO}=\text{ distance fallen in time }\mathrm{t}=\frac{1}{2}{\mathrm{gt}}^2 \\ \text{ At time ' }t\text{ ', }u=-\left[h-\frac{1}{2}g{t}^2\right] \\ \mathrm{V}=?{\mu }_1=1,{\mu }_2=\mu ,\mathrm{R}=+\mathrm{R} \\ \frac{\mathrm{\mu }}{\mathrm{V}}-\frac{1}{-\left(\mathrm{h}-\frac{1}{2}{\mathrm{gt}}^2\right)}=\frac{\mathrm{\mu }-1}{\mathrm{R}} \\ \Rightarrow \mathrm{V}=\frac{\mathrm{\mu R}\left(\mathrm{h}-\frac{1}{2}{\mathrm{gt}}^2\right)}{(\mathrm{\mu }-1)\left(\mathrm{h}-\frac{1}{2}{\mathrm{gt}}^2\right)-\mathrm{R}} \end{gathered}$$

= Image distance from ‘P’ at time ‘t’ 

Image velocity at time ‘t’ is 
$$\begin{gathered} {\mathrm{V}}_{\mathrm{I}}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left[\frac{\mathrm{\mu R}\left(\mathrm{h}-\frac{1}{2}{\mathrm{gt}}^2\right)}{(\mathrm{\mu }-1)\left(\mathrm{h}-\frac{1}{2}{\mathrm{gt}}^2\right)-\mathrm{R}}\right] \\ =\frac{\left[\left[(\mathrm{\mu }-1)\left(\mathrm{h}-\frac{1}{2}{\mathrm{gt}}^2\right)-\mathrm{R}\right](-\mathrm{gt\mu R})\right]-\left[\mathrm{\mu R}\left(\mathrm{h}-\frac{1}{2}{\mathrm{gt}}^2\right)\left[\right(\mathrm{\mu }-1\left)\right(-\mathrm{gt}\left)\right]\right]}{{\left[(\mathrm{\mu }-1)\left(\mathrm{h}-\frac{1}{2}{\mathrm{gt}}^2\right)-\mathrm{R}\right]}^2} \\ =\frac{{\mathrm{\mu R}}^2\mathrm{gt}}{{\left[(\mathrm{\mu }-1)\left(\mathrm{h}-\frac{1}{2}{\mathrm{gt}}^2\right)-\mathrm{R}\right]}^2} \end{gathered}$$