= Solution
For the stated standard configuration, the <Lorentz transformation> is
$$
\boxed{t'=\gamma(t-vx/c^2),\quad x'=\gamma(x-vt),\quad y'=y,\quad z'=z,\quad
\gamma=(1-v^2/c^2)^{-1/2}.}
$$
Differentiate along the moving object's worldline and divide $dx'$ by $dt'$ to obtain the <velocity-addition formula>:
$$
\boxed{u_x'=\frac{u_x-v}{1-u_xv/c^2}.}
$$
For a receding source, two emissions separated by proper time $1/f'$ occur $\gamma/f'$ apart in S. The second signal travels an additional distance $v\gamma/f'$, so its reception interval is $\gamma(1+v/c)/f'$. Hence the longitudinal <relativistic Doppler effect> is
$$
\boxed{f=\frac{f'}{\gamma(1+v/c)}=f'\sqrt{\frac{1-v/c}{1+v/c}}.}
$$
The inverse velocity transformation for the forward throw gives
$$
\boxed{u=\frac{c/2+c/7}{1+1/14}=\frac35c.}
$$
The ball approaches the stationary receiver, so its forward light is blueshifted. The received wavelength is $400\sqrt{(1-3/5)/(1+3/5)}=\boxed{200\,\mathrm{nm}}$, in the <ultraviolet>. \b[The stated monochromatic emission is outside the visible band.] Other reflected or emitted frequencies have not been specified. <University Physics Volume 3 by Samuel J. Ling and others>[§5.7, longitudinal light Doppler shift and approach/recession signs]{ref}
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