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.