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WAEC 2016 Physics · Question 25

At a birthday party, the celebrant pops a corked fruit wine. If the cork shoots out of the bottle at an angle…

Question

At a birthday party, the celebrant pops a corked fruit wine. If the cork shoots out of the bottle at an angle of 40o to the horizontal and travels a horizontal distance of 4.50m in 1.25s. Calculate the initial speed of the cork.

  1. A.4.2ms−1ms^{-1}
  2. B.4.7ms−1ms^{-1}
  3. C.5.6ms−1ms^{-1}
  4. D.7.1ms−1ms^{-1}

Answer: B. 4.7ms−1ms^{-1}

Explanation written with AI and checked against the answer key

Analyze the question: This is projectile motion. The cork leaves at 40° to the horizontal and covers a horizontal distance (range) of 4.50 m in 1.25 s. We need the initial speed. Concept: The horizontal component of velocity stays constant, so: Horizontal distance = u cosθ × t Solution: 4.50 = u cos 40° × 1.25 u cos 40° = 4.50 / 1.25 = 3.6 m/s cos 40° = 0.7660 u = 3.6 / 0.7660 ≈ 4.70 m/s Check with the other formulas: If we assume the cork lands at the same height it started from, the time of flight would be T = 2u sinθ / g. With u = 4.7, T = 2(4.7)(0.6428)/10 ≈ 0.60 s, which does not match 1.25 s. So the data are not fully consistent with a level landing (the cork may have landed on a lower surface, such as the floor). The question only asks for the speed from the horizontal data, so the horizontal-motion method is the intended approach, and it matches option B. Answer: u ≈ 4.7 m/s, which is option B. Tip: When a projectile question gives a horizontal distance and a time, go straight to x = (u cosθ)t. Don't use the range formula R = u² sin2θ / g unless the time is not given and the landing height equals the launch height.

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