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The alpha particle has electric charge $2e$ and acquires kinetic energy $K=2e(20\mathrm{kV})=40\mathrm{keV}$. For Coulomb head-on closest approach with target recoil, the initially stationary beryllium nucleus is free to recoil. Approximating the nuclear masses by $4u$ and $9u$, the initial relative-motion kinetic energy isAt closest separation the relative radial velocity vanishes, although the centre of mass continues moving. Conservation of energy for the relative motion givesUsing $e^2/(4\pi {\epsilon }_0)=1.43996\mathrm{MeV}\mathrm{fm}$ gives $r_{\min }\simeq 416\mathrm{fm}$. If the nucleus were externally held fixed, the relevant energy would be $K$, yielding $288\mathrm{fm}$ instead. Initially stationary is not the same as constrained stationary; the given proton and neutron counts supply the recoil correction. Nuclear sizes and strong interactions are negligible at this separation.

Choose the positive axis from the approaching source towards the observer. The photon that reaches the observer is emitted forward, with source-frame four-momentum $(E/c,E/c,0,0)$. Boosting to the observer frame gives $E_{\mathrm{obs}}=\gamma E(1+\beta )$, where $\beta =v/c$. Since photon energy is $h_{\mathrm{P}}\nu$,This is the approaching-source relativistic Doppler effect. A photon emitted in a general source-frame direction would instead have the factor $\gamma (1+\beta \cos {\theta }_{\ast })$; forward emission is the geometry of the photon directed to the observer here.

If $S^{\prime }$ moves at $+v$ relative to $S$, the Lorentz transformations areA rod at rest in $S$ has length $L$. To measure it in $S^{\prime }$, select endpoint events simultaneous there: $\Delta t^{\prime }=0$, so $\Delta t=vL/c^2$ and $\Delta x^{\prime }=\gamma (L-v\Delta t)=L/\gamma$. This is length contraction.

Now let a clock at rest in $S^{\prime }$ cross the two rod endpoints. These are different events, with $\Delta x=L$, $\Delta t=L/v$ and $\Delta x^{\prime }=0$. Their transformed time interval isThusThe moving clock measures the proper time $\Delta t^{\prime }$. The rod's contracted length divided by its speed in that clock's frame gives exactly the same crossing time; time dilation and length contraction are consistent consequences of the same transformation. Measuring a length and measuring a traversal interval require different simultaneity conditions.

For maximum projectile range from an elevated launch, let $X$ be horizontal projectile range from the tower and $\alpha$ the launch angle above horizontal. The projectile motion trajectory isPut $z=\tan \alpha$. For a fixed $X$, its largest possible $y$ occurs at $z=u^2/(gX)$; the envelope is $y_{\max }=h+u^2/(2g)-g{X}^2/(2u^2)$. The furthest attainable ground point satisfies $y_{\max }=0$, soThe optimum angle is less than $45^{\circ }$ for $h>0$ and approaches $45^{\circ }$ as $h\to 0$. The result is the maximum horizontal distance, using uniform gravity and no drag.