WAEC Further Mathematics 2025 Paper

Concept:Use the equation of motion under constant acceleration: $s = ut + \frac{1}{2}at^2$.Explanation:Take upward direction as positive.Initial height is $45$ m above the ground, so final displacement to the ground is $s = -45$ m.Initial velocity $u = 40$ m/s upward.Acceleration due to gravity acts downward: $a = -g = -10$ m/s$^2$.Substitute into $s = ut + \frac{1}{2}at^2$:$$-45 = 40t + \frac{1}{2}(-10)t^2$$Simplify:$$-45 = 40t - 5t^2$$Rearrange into standard quadratic form:$$5t^2 - 40t - 45 = 0$$Divide through by $5$:$$t^2 - 8t - 9 = 0$$Solve using the quadratic formula:$$t = \frac{8 \pm \sqrt{(-8)^2 - 4(1)(-9)}}{2} = \frac{8 \pm \sqrt{100}}{2}$$So $t = \frac{8 + 10}{2} = 9$ s or $t = \frac{8 - 10}{2} = -1$ s.Discard $t = -1$ s because time cannot be negative.Therefore, the particle hits the ground after $9$ seconds.Answer:The time taken to hit the ground is $9$ seconds, which is Option A.