A ball is held at a height h above the surface of a heavy...

The distance of the ball from the pole of the spherical surface at time $t$ is given by $y = h - \dfrac{1}{2}gt^2$.

Using the sign convention, the object distance is u = -y = - \left(h -\dfrac{1}{2}gt^2\right).

The velocity of the object is $v_o =$\dfrac{du}{dt} =$gt$.

The formula for refraction at a single spherical surface is:

\dfrac{\mu_2}{v} - \dfrac{\mu_1}{u} = \dfrac{\mu_2 - \mu_1}{R}

Substituting $\mu_1 = 1$ and $\mu_2 = \mu$, we get:

\dfrac{\mu}{v} - \dfrac{1}{u} = \dfrac{\mu - 1}{R}

Differentiating this equation with respect to time $t$ yields:

-\dfrac{\mu}{v^2} \dfrac{dv}{dt} + \dfrac{1}{u^2} \dfrac{du}{dt} =$0$

\dfrac{dv}{dt} = \dfrac{v^2}{\mu u^2} \dfrac{du}{dt}

From the refraction formula, we can find the ratio \dfrac{v}{u}:

\dfrac{\mu}{v} = \dfrac{R + u(\mu - 1)}{u R}

\dfrac{v}{u} = \dfrac{\mu R}{R + u(\mu - 1)}

Squaring this ratio and dividing by $\mu$ gives:

\dfrac{v^2}{\mu u^2} = \dfrac{\mu R^2}{[R + u(\mu - 1)]^2}

Substituting $u = - \left(h - \dfrac{1}{2}gt^2\right)$ into the denominator:

R + u(\mu - 1) = R - (\mu - 1) \left(h -\dfrac{1}{2}gt^2\right)

Since this term is squared, we can reverse the sign inside the bracket:

[R + u(\mu - 1)]^2 = \left[(\mu - 1)\left(h -\dfrac{1}{2}gt^2\right) - R\right]^2

The speed of the image is the magnitude of \dfrac{dv}{dt}:

$v_i =$\dfrac{\mu R^2}{\left[(\mu - 1)\left(h - \dfrac{1}{2}gt^2\right) - R\right]^2} ($gt)$

$v_i = \dfrac{\mu R^2 gt}{\left[(\mu - 1)\left(h - \dfrac{1}{2}gt^2\right) - R\right]^2}$