Special Relativity - Finding Velocity

[Insolite]

A spaceship is measured to be 50m long in it's own rest frame takes ##1.50 \times 10^{-6} s## to pass overhead, as measured by an observer on earth. What is its speed relative to earth?

My attempt at the solution involves the use of the equation for length contraction, ##l = l_{0}\sqrt{1 - \frac{u^{2}}{c^{2}}}##. I have calculated an answer that seems reasonable, but I don't know if the approach that I've taken is correct.

Solution:

##l = l_{0}\sqrt{1 - \frac{u^{2}}{c^{2}}}##, where ##l_{0} = 50m##, ##u = \frac{l}{t}## with respect to earth.
##l = l_{0}\sqrt{1 - \frac{u^{2}}{c^{2}}}##

##l^{2} = l_{0}^{2}(1 - \frac{u^{2}}{c^{2}})##

##\frac{1}{l_{0}^{2}}l^{2} + \frac{u^{2}}{c^{2}} = 1##

##\frac{l^{2}}{l_{0}^{2}} + \frac{l^{2}}{(tc)^{2}} = 1##

##\frac{l^{2}c^{2}t^{2} + l_{0}^{2}l^{2}}{l_{0}^{2}t^{2}c^{2}} = 1##

##l^{2}t^{2}c^{2} + l^{2}l_{0}^{2} = l_{0}^{2}t^{2}c^{2}##

Substituting in the known values of ##l_{0} = 50m## and ##t = 1.50 \times 10^{-6}##:

##l = 49.694m##

Therefore, the spaceship travels this length (which is its length with respect to the earth) in ##1.5 \times 10^{-6} s## with respect to the earth:

Speed = ##\frac{49.694m}{1.5 \times 10^{-6} s}##

## = 3.31 \times 10^{7}ms^{-1}##

If this is in fact correct, are there any other methods that I may have used to arrive at the same answer? (Those using very basic special relativity concepts, including time dilation and Lorentz transformations).

Thanks for any help and advise.

 

[mfb]

Looks right but complicated.
As seen from earth, the spaceship has a speed of ##\beta c## and a length contraction factor ##\gamma##, so we get ##t = \frac{l_0}{\beta \gamma}## or
$$t \beta c = l_0 \sqrt{1-\beta^2}$$
$$\frac{\beta}{\sqrt{1-\beta^2}} = \frac{l_0}{tc}$$
The right side is a known value (0.111) and you can directly solve the equation for β.

The result is the same.