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**Problem:** A spaceship travels past Earth at a speed of 0.8 c. An astronaut on the ship measures a time interval of 5 seconds between two lightning strikes that occur at the front and back of the ship. What is the time interval measured by an observer on Earth? **Solution Steps:** 1. Identify the speed: v = 0.8 c. 2. Compute the Lorentz factor: \[\gamma = \frac{1}{\sqrt{1 - v^{2}/c^{2}}} = \frac{1}{\sqrt{1 - (0.8)^{2}}} = \frac{1}{\sqrt{1 - 0.64}} = \frac{1}{\sqrt{0.36}} = \frac{1}{0.6} = 1.667.\] 3. The proper time (time measured in the moving frame) is \(\Delta t_{0}=5\) s. 4. Time dilation formula: \(\Delta t = \gamma \Delta t_{0}\). 5. Plug values: \(\Delta t = 1.667 \times 5\text{ s} = 8.33\text{ s}\). **Answer:** An observer on Earth sees the interval as about **8.33 seconds**. **Interpretation:** Because the spaceship is moving fast, the Earth’s clock records a longer time than the astronaut’s clock. **Length‑contraction check (optional):** If the ship’s proper length is 300 m, the length seen from Earth would be L = L₀/γ = 300 m / 1.667 ≈ 180 m.