[Physics] Dropping a ball in a train moving close to the speed of light

special-relativity

Suppose a train is moving very close to the speed of light, say 0.999c relative to a stationary observer on Earth.

Now a stationary observer on Earth will observe clocks on the train to tick slower than usual.

Now suppose a boy within the train drops a ball.

The stationary observer measures how long it takes for the ball to hit the ground. Will the result simply be $\sqrt{\frac{2s}{g}}$ or will the result be a greater number due to time dilation?

Rely on the invariance of the interval $\tau$ between the ball being dropped and the ball hitting the floor. Also using $c=1$ units to keep the typing down.

The moving frame is un-primed and the stationary frame primed.

$$t^2 - h^2 = \tau^2 = {t'}^2 - [{h'}^2 + (vt')^2] \,.$$

Because the velocity $v$ is horizontal both parties measure the same $h$, allowing us to write

\begin{align*} {t'}^2 - [h^2 + (vt')^2] - \tau^2 &= 0 \\ {t'}^2(1 - v^2) - h^2 - \tau^2 &= 0 \\ {t'}^2(1 - v^2) - h^2 - (t^2 - h^2) &= 0 \\ {t'}^2(1 - v^2) - t^2 &= 0 \\ t' = \frac{t}{\sqrt{1 - v^2}} \,, \end{align*}

so I owe Mew an apology for my comments in the earlier thread, but it does not violate the equivalence principle because we didn't use any information about the value of $h$ in the above work, so it works fine when $h=0$ as in the elevator case.