1. **Problem Statement:** We need to determine if the sailboat with a draft of 2 m can safely exit the harbour at 6:30 p.m. given the tide heights at 2 p.m. (high tide at 10 m) and 8:15 p.m. (low tide at 1.2 m). 2. **Understanding the Tide Model:** The tide height varies sinusoidally between a high point and a low point. We can model the tide height $h(t)$ as a sinusoidal function: $$h(t) = A \sin\left(B(t - C)\right) + D$$ where: - $A$ is the amplitude (half the difference between high and low tide), - $B$ relates to the period of the tide, - $C$ is the horizontal shift (time of high tide), - $D$ is the vertical shift (average of high and low tide), - $t$ is the time in hours. 3. **Calculate Parameters:** - High tide at $t=14$ (2 p.m.) with height $10$ m. - Low tide at $t=20.25$ (8:15 p.m.) with height $1.2$ m. Amplitude: $$A = \frac{10 - 1.2}{2} = \frac{8.8}{2} = 4.4$$ Vertical shift: $$D = \frac{10 + 1.2}{2} = \frac{11.2}{2} = 5.6$$ Period $T$ is twice the time between high and low tide: $$T = 2 \times (20.25 - 14) = 2 \times 6.25 = 12.5 \text{ hours}$$ Angular frequency: $$B = \frac{2\pi}{T} = \frac{2\pi}{12.5} = 0.5027$$ Horizontal shift $C$ is the time of high tide: $$C = 14$$ 4. **Tide height function:** $$h(t) = 4.4 \sin\left(0.5027 (t - 14)\right) + 5.6$$ 5. **Calculate tide height at 6:30 p.m. ($t=18.5$):** $$h(18.5) = 4.4 \sin\left(0.5027 (18.5 - 14)\right) + 5.6 = 4.4 \sin(0.5027 \times 4.5) + 5.6$$ Calculate inside sine: $$0.5027 \times 4.5 = 2.262$$ Calculate sine: $$\sin(2.262) \approx 0.770$$ Calculate height: $$h(18.5) = 4.4 \times 0.770 + 5.6 = 3.388 + 5.6 = 8.988$$ 6. **Conclusion:** The tide height at 6:30 p.m. is approximately $8.99$ m, which is well above the sailboat's draft of 2 m. Therefore, the captain can safely exit the harbour at 6:30 p.m. **Final answer:** Yes, the captain can safely exit the harbour at 6:30 p.m. because the water depth is about 8.99 m, which is greater than the 2 m draft of the sailboat.