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WAEC 2015 Physics · Question 23

A bullet fired vertically upwards reaches a height of 500m. Neglecting air resistance, calculate the magnitude…

Question

A bullet fired vertically upwards reaches a height of 500m. Neglecting air resistance, calculate the magnitude of the initial velocity of the bullet. [g = 10ms−2ms^{-2}]

  1. A.500.0ms−1ms^{-1}
  2. B.100.0ms−1ms^{-1}
  3. C.70.0ms−1ms^{-1}
  4. D.50.0ms−1ms^{-1}

Answer: B. 100.0ms−1ms^{-1}

Explanation written with AI and checked against the answer key

Analyze the question: The bullet goes straight up and stops momentarily at the maximum height (500 m), where its final velocity v = 0. We need the initial velocity u. Concept: For motion under gravity, use the equation v² = u² − 2gh (g is negative because the bullet is moving against gravity.) Solution: Given: v = 0, h = 500 m, g = 10 m/s² 0 = u² − 2(10)(500) u² = 10,000 u = √10,000 = 100 m/s Check: Maximum height H = u²/2g = (100)²/(2×10) = 10,000/20 = 500 m ✓ Tip: For any "maximum height" question, remember that v = 0 at the top, so u² = 2gH. Square-root numbers like 10,000 quickly: √10,000 = 100. If you get stuck, you can also test the options in H = u²/2g. Option A would give 12,500 m, which is far too high, while option B gives exactly 500 m.

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