In 2012, professional sky diver and crazy person Felix Baumgartner jumped to Ear

Question

In 2012, professional sky diver and crazy person Felix Baumgartner jumped to Earth from a capsule suspended by a balloon from a height of about 39 kilomters. During this jump he set the record for the fastest speed of free fall, becoming the first human to break the sound barrier outside of a vehicle. He took very little convincing to try my next high-speed experiment. We strapped a "warp speed" rocket onto Felix's back and shot him into outer space. Because he doesn't need to worry about air resistance in space, he decided to hold his arms out from his sides at an angle theta = 45 degree with respect to his body and pretended that they were his "wings" (he is sponsored by Red Bull, after all). To a stationary observer on the Moon (we will consider the Moon stationary for our purposes), at what angle will it appear that Felix is holding out his arms with respect to his body when he goes speeding by head-first in the -x direction at speed upsilon = 0.6c?

Explanation / Answer

According to the question

We know that

Assume v is the radial velocity towards the moon in Flex referance frame towards the moon. Also in Flex referenace frame, he makes theta angle from his side. If X is his side length & Y distance from his side

then,

tan theta = Y/X

Here

In Flex ref. frame Vy= 0 & Vx= v

So,

In Moon referance frame Xm= X/0.6while Ym= Y

where

y= 1/(1-v^2/c^2)^1/2

tan 45 m  =Ym/Xm =(Y.0.6)/X =0.6.tan 45

or 45 m= tan^-1 (y.tan theta), where 0.6 is as above.

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