Answer
a. 15 meters.
b. 42 minutes.
Work Step by Step
a. Observers on Earth measure the ship as shorter than the rest length. Use equation 26–3a.
$$\mathcal{l}=\mathcal{l}_0\sqrt{1-v^2/c^2}=(23m)\sqrt{1-(0.75)^2}= 15m$$
b. Observers on Earth measure the time interval as longer than the proper time. Use equation 26–1a.
$$\Delta t = \frac{\Delta t_0}{\sqrt{1-v^2/c^2}}= \frac{28min}{\sqrt{1-(0.75)^2}}=42min$$