Answer
The painting will be 0.94 meters long, and 0.89 meters high.
Work Step by Step
We are told that the ship travels at a speed of 0.90c at an angle of $30^{\circ}$ above the horizontal. Resolve the motion in two components.
In the horizontal direction, the velocity is $(0.90c) cos 30^{\circ}=0.779c$.
Use equation 26–3a to find the contracted length.
$$\mathcal{l}=\mathcal{l}_0\sqrt{1-v^2/c^2}=1.50m\sqrt{1-0.779^2}=0.94m$$
In the vertical direction, the velocity is $(0.90c) sin 30^{\circ}=0.450c$.
Use equation 26–3a to find the contracted height.
$$h=h_0\sqrt{1-v^2/c^2}=1.00m\sqrt{1-0.450^2}=0.89m$$
