WAEC Further Mathematics 2025 Paper

The motion of the particle can be described using the equation of displacement under constant acceleration:$s = ut + \frac{1}{2}at^2$where:$s$ is the displacement (here, $s = -45$ m, taking upward as positive and the ground as the final position),$u = 40$ m/s (initial velocity upward), $a = -g = -10$ m/s² (acceleration due to gravity, approximating $g = 10$ m/s²),- $t$ is the time.$-45 = 40t + \frac{1}{2}(-10)t^2$$-45 = 40t - 5t^2$Rearrange into standard quadratic form:$5t^2 - 40t - 45 = 0$Divide through by 5:$t^2 - 8t - 9 = 0$Solve using the quadratic formula:$t = \frac{8 \pm \sqrt{(-8)^2 - 4(1)(-9)}}{2} = \frac{8 \pm \sqrt{64 + 36}}{2} = \frac{8 \pm \sqrt{100}}{2} = \frac{8 \pm 10}{2}$The solutions are:$t = \frac{8 + 10}{2} = 9 \text{ s}$$t = \frac{8 - 10}{2} = -1 \text{ s (discard, as time cannot be negative).}$Thus, the time taken to hit the ground is 9 seconds.