Answer
(a) Observers on the spaceship would also measure a height of $0.50~m$.
(b) The width according to others on the spaceship is $2.1~m$
Work Step by Step
(a) Since the height is perpendicular to the direction of motion, there is no height contraction. Observers on the spaceship would also measure a height of $0.50~m$.
(b) Let $W = 0.50~m$. We can find the width $W_0$ measured by the observer on the spaceship:
$W = W_0~\sqrt{1-\frac{v^2}{c^2}}$
$W_0 = \frac{W}{\sqrt{1-\frac{v^2}{c^2}}}$
$W_0 = \frac{0.50~m}{\sqrt{1-\frac{(0.97~c)^2}{c^2}}}$
$W_0 = 2.1~m$
The width according to others on the spaceship is $2.1~m$
