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NECO 2003 Physics · Question 16

A stone is released from a height of 80m above the ground. Calculate its velocity just before it strikes the ground. (g = 10ms−2^{-2})

Options

  1. A.25ms−1ms^{-1}
  2. B.40 ms−1ms^{-1}
  3. C.55 ms−1ms^{-1}
  4. D.65ms−1ms^{-1}
  5. E.80 ms−1ms^{-1}

Answer: B. 40 ms−1ms^{-1}

Explanation written with AI and checked against the answer key

Analyze the question: A stone is released (so initial velocity u = 0) from a height h = 80 m. We need its speed just before it hits the ground, with g = 10 m/s². Concept: This is free fall under gravity, so we use the equations of motion with a = g: v² = u² + 2gh (Alternatively, use energy conservation: mgh = ½mv², which gives the same result.) Solution: Given: u = 0, g = 10 m/s², h = 80 m v² = 0² + 2(10)(80) v² = 1600 v = √1600 = 40 m/s Cross-check with time: h = ½gt², so 80 = 5t², giving t² = 16 and t = 4 s. v = u + gt = 0 + 10(4) = 40 m/s ✓ Answer: B (40 m/s) Tip: For "dropped from rest" questions, go straight to v = √(2gh). Mass does not matter. It is also helpful to remember that with g = 10, a drop of 80 m gives 4 s and 40 m/s. Be careful not to pick 80 (option E), which comes from mistakenly using v = h or forgetting the square root.

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