Question

A spherical UFO with a radius of 67 m flies past Earth with a velocity of `2.9*10^8` `ms^-1`. What is its height and width as measured by observers on the Earth?

Answer 1

Height = 134m
Width `~~34`m

Explanation:

For this scenario we will assume that the observers and the UFO are parallel, that the observers are at rest with only the UFO moving, and that the UFO is flying in a horizontal direction to the observers eyes, who are standing upright.

The height remains unchanged because the length contraction is only being measured in one plane, which happens to be the plane on which the width of the UFO is located.

To measure the contraction of the width, there is a very nifty formula that solves that does it quickly, and it looks like this:
`L=L_0sqrt(1-v^2/c^2`,
where `L_0` is the proper length of the object in its rest frame, `L` is the contracted length (observed by the observer in relative motion with respect to the object), `v` is the relative velocity between object and observer, and `c` is the speed of light.

From the question asked it becomes clear that we have all the necessary value, and so we simply plug them into the formula, like so:

`L=134sqrt(1-(2.9*10^8)^2/(2.997*10^8)^2`
`L=134sqrt(1-.93574)`
`L=134sqrt0.06426`
`L~~33.9687`m

Notice how I doubled the radius so as to get the proper length, or width, of the object. Therefore the width of the UFO as seen by observers on Earth will be roughly 34m.

https://en.wikipedia.org/wiki/Length_contraction#Experimental_verifications

I hope I helped!