A missile is launched with a speed of 75 ms^ -1 at an angle of 22° above the surface of a warship. Find the…

Analyze the question: This is a projectile motion problem. We're given the launch speed (u = 75 m/s), the launch angle (θ = 22°), and g = 10 m/s². We need the horizontal range R.

Concept: For a projectile landing at the same height it was launched from:

R = u² sin(2θ) / g

Solution:

1. u² = 75² = 5625
2. 2θ = 44°
3. sin 44° ≈ 0.6947
4. R = (5625 × 0.6947) / 10
5. R = 3907.7 / 10 ≈ 390.8 m

So R ≈ 391 m.

Answer: Option C (391 m).

Tip: Remember that range is maximum at 45° because sin 90° = 1. Here 2θ = 44° is very close to 90°, so sin 44° is close to 0.7. That means R should be close to 0.7 × 562.5 ≈ 394 m, which lets you quickly eliminate options A, B and D. (Option A also has no unit, which is a hint it's a weak distractor.)

Also remember to set your calculator to degree mode when evaluating sin 44°.