WAEC Physics 2011 Paper

Concept: This is a horizontal projectile problem. The coin's time of fall depends only on the bench height and gravity, while its horizontal distance depends on its horizontal velocity.Explanation:Vertical motion starts with zero vertical velocity, so $h = \frac{1}{2}gt^{2}$. Here $h = 1.5\,\mathrm{m}$ and $g = 10\,\mathrm{m\,s^{-2}}$. Time to fall: $$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 1.5}{10}} = \sqrt{0.3} = 0.5477\,\mathrm{s}$$ Horizontal distance from the foot of the bench: $$R = u t = 15.0 \times 0.5477 = 8.22\,\mathrm{m}$$ Using the combined formula: $$R = u\sqrt{\frac{2h}{g}} = 15.0\sqrt{\frac{2 \times 1.5}{10}} = 8.22\,\mathrm{m}$$Answer: The distance is $8.22\,\mathrm{m}$, which is option C.