Discussion
A particle of mass \(m\) falling vertically under gravity. Then the Lagrangian L of the particle is given by :
Answer: B
With \(z\) measured downward along the line of fall, the Lagrangian is \(L = \frac{1}{2}m\dot{z}^{2} + mgz\).
The motion is one-dimensional, so a single coordinate \(z\) suffices, which rules out the two-coordinate forms in \(x\) and \(y\). For a falling particle the usual textbook choice measures \(z\) downward from the point of release. Gravity then does positive work as \(z\) increases, the potential energy is \(V = -mgz\), and
\[ \begin{aligned} L &= T - V = \frac{1}{2}m\dot{z}^{2} + mgz \\ m\ddot{z} &= mg \end{aligned} \]
which is free fall with acceleration \(g\) in the direction of increasing \(z\), as it should be.
The form \(\frac{1}{2}m\dot{z}^{2} - mgz\) is the same physics with \(z\) measured upward; the paper's key takes the downward convention, which is the natural one for a body described as falling.
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