Linear Motion & Projectile Motion — A/L Physics

How do I practise Linear Motion & Projectile Motion?

1 question with a worked solution

1. A projectile is launched from ground level with speed u at an elevation angle θ to the horizontal. If air resistance is negligible, what is the maximum height H achieved by the projectile?
   1. A) H = (u² cos² θ) / (2g)
   2. B) H = (u² sin² θ) / (2g)
   3. C) H = (u² sin 2θ) / g
   4. D) H = (u sin θ) / g
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   Correct answer: B) H = (u² sin² θ) / (2g)

   At the apex of projectile motion, the vertical velocity component becomes zero: v_y = 0. Using v_y² = u_y² - 2gH with u_y = u sin θ: 0 = (u sin θ)² - 2gH, giving H = (u² sin² θ) / (2g).

   Show worked solution

   Step 1: Vertical motion equation: v_y² = u_y² - 2gH. Step 2: At maximum height, v_y = 0 and u_y = u sin θ. Step 3: 0 = u² sin² θ - 2gH => H = (u² sin² θ) / (2g). Answer: H = (u² sin² θ) / (2g).
   Modelled on: the A/L Physics examination pattern

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Where does Linear Motion & Projectile Motion fit in the syllabus?

Unit 2.1 · Year 1 · Term 1