An aeroplane flies over a rectangular field of dimensions 100 m ×...

Let the proper length of the field be $L_0 = 100 \text{ m}$ and the proper width be $W_0 = 50 \text{ m}$.

The aeroplane moves parallel to the length of the field with speed $v$. According to the theory of relativity, the length of the field observed from the aeroplane undergoes length contraction, while the width remains unchanged.

The observed length is given by $L = L_0 \sqrt{1 - \dfrac{v^2}{c^2}}$.

The observed width is $W = W_0 = 50 \text{ m}$.

For the field to appear as a square in the aeroplane's frame of reference, the observed length must be equal to the observed width:

$L = W$

$100 \sqrt{1 - \dfrac{v^2}{c^2}} = 50$

$\sqrt{1 - \dfrac{v^2}{c^2}} = \dfrac{1}{2}$

Squaring both sides:

$1 - \dfrac{v^2}{c^2} = \dfrac{1}{4}$

$\dfrac{v^2}{c^2} = \dfrac{3}{4}$

$v = \dfrac{\sqrt{3}}{2} c$