Given that an observer on Earth with acceleration due to gravity 9.8 m/s^2 has an indestructible frictionless gyroscope and a means by which to safely accelerate it to a rotational velocity of (some relativistic velocity, we'll say).85 c, how long will the gyroscope take to fall in a vacuum (I think the problem may have said neglecting air resistance) relative to the observer if dropped from a height of 15 meters? So to sum it up, g=9.8 m/s^2, and this 'indestructible' gyroscope with 'negligible friction' (Yes it's a crazy question) is dropped from a height of 15 meters through a vacuum while spinning at .85 c. How long will it take to fall relative to an outside observer?

Lorentz equations are necessary for sure, but it's not the calculations that really stumped me, it's the theory behind it. Part of me says (classically) it will fall just like anything else. If I'm not mistaken though special relativity applies to the gyroscope because it's moving uniformly. So would the gyroscope in effect fall "slower" to the observer due to time dilation occurring for an observer "on the gyroscope" such as an ant? As I said, I assumed it was a trick and wrote it off as simple mechanics. I don't even think it was a fair question to ask, mainly because it's slowly ruining my summer. Can anyone help me out?