WAEC Physics 2014 Paper

Concept:Use the equation of motion for constant acceleration, including the initial height above the ground.Explanation:Take the point of projection as the origin and upward direction as positive.The ground is $2\ \text{m}$ below the origin, so the displacement until landing is $s = -2\ \text{m}$.Initial velocity is $u = +10\ \text{m s}^{-1}$ and acceleration is $a = -10\ \text{m s}^{-2}$.Using $s = ut + \frac{1}{2}at^2$:$$-2 = 10t - \frac{1}{2}(10)t^2$$$$-2 = 10t - 5t^2$$Rearranging gives:$$5t^2 - 10t - 2 = 0$$Solve using the quadratic formula:$$t = \frac{10 \pm \sqrt{100 + 40}}{10} = \frac{10 \pm \sqrt{140}}{10}$$Taking the positive root:$$t = 1 + \frac{\sqrt{35}}{5} \approx 2.18\ \text{s}$$Answer:The total time taken is approximately $2.18\ \text{s}$, so the correct option is C.