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WAEC 2013 Physics · Question 20

A body of mass 20g projected vertically upwards in vacuum returns to the point of projection after 1.2s. [g = 10ms−2^{-2}]. Calculate the speed of projection

Options

  1. A.1.2ms−1ms^{-1}
  2. B.6.0ms−1ms^{-1}
  3. C.12.0ms−1ms^{-1}
  4. D.0.6ms−1ms^{-1}

Answer: B. 6.0ms−1ms^{-1}

Explanation written with AI and checked against the answer key

Analyze the question: A body is thrown straight up and returns to its starting point after a total time of flight T = 1.2 s. We need the speed of projection u. The mass (20 g) is not needed, because in a vacuum there is no air resistance and acceleration is always g. Concept: For motion under gravity, the time of flight is T = 2u/g This works because the body takes the same time to rise as to fall. The time up is T/2, and at the top the velocity is zero. Solution: Time to reach the maximum height: t = T/2 = 1.2/2 = 0.6 s Using v = u − gt, with v = 0 at the top: 0 = u − (10)(0.6) u = 6.0 m/s Check with the displacement formula (s = 0 on return): s = ut − ½gt² 0 = u(1.2) − 5(1.2)² u(1.2) = 5(1.44) = 7.2 u = 6.0 m/s ✓ Answer: The speed of projection is 6.0 m/s, which is option B. Tip: In "returns to the point of projection" questions, always halve the total time to get the time of ascent. Then use u = g × (T/2). Ignore the mass, as it is a distractor here.

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