1. **State the problem:** We have a golf ball hit from the top of a cliff. The height of the ball above the ground at time $t$ seconds is given by the function $$h(t) = 40 + 10t - 5t^2.$$ We need to find: (i) The height of the cliff. (ii) The height of the ball after 1.5 seconds. (iii) The maximum height of the ball. (iv) The time the ball is in the air before hitting the ground. (v) The distance between the ball and the top edge of the cliff when it landed, given it landed 240 m from the foot of the cliff. 2. **Formula and rules:** The height function is a quadratic equation in the form $$h(t) = at^2 + bt + c$$ where $a = -5$, $b = 10$, and $c = 40$. The graph is a parabola opening downward because $a < 0$. 3. **(i) Height of the cliff:** This is the height at time $t=0$. $$h(0) = 40 + 10 \times 0 - 5 \times 0^2 = 40$$ So, the cliff is 40 metres high. 4. **(ii) Height after 1.5 seconds:** Substitute $t=1.5$ into $h(t)$: $$h(1.5) = 40 + 10 \times 1.5 - 5 \times (1.5)^2 = 40 + 15 - 5 \times 2.25 = 40 + 15 - 11.25 = 43.75$$ The ball is 43.75 metres above the ground after 1.5 seconds. 5. **(iii) Maximum height:** The vertex of the parabola gives the maximum height. The time at vertex is $$t = -\frac{b}{2a} = -\frac{10}{2 \times (-5)} = -\frac{10}{-10} = 1$$ Substitute $t=1$ into $h(t)$: $$h(1) = 40 + 10 \times 1 - 5 \times 1^2 = 40 + 10 - 5 = 45$$ Maximum height is 45 metres. 6. **(iv) Time ball hits the ground:** The ball hits the ground when $h(t) = 0$. Solve: $$0 = 40 + 10t - 5t^2$$ Rewrite: $$-5t^2 + 10t + 40 = 0$$ Divide both sides by $-5$: $$\cancel{-5}t^2 - \cancel{10}t - \cancel{40} = 0 \Rightarrow t^2 - 2t - 8 = 0$$ Use quadratic formula: $$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4 \times 1 \times (-8)}}{2} = \frac{2 \pm \sqrt{4 + 32}}{2} = \frac{2 \pm \sqrt{36}}{2}$$ $$t = \frac{2 \pm 6}{2}$$ Two solutions: $$t = \frac{2 + 6}{2} = 4$$ $$t = \frac{2 - 6}{2} = -2$$ (discard negative time) So, the ball hits the ground after 4 seconds. 7. **(v) Distance between ball and cliff top when landed:** The ball landed 240 m horizontally from the foot of the cliff. The vertical distance is 0 (on ground), cliff top is 40 m high. Use Pythagoras theorem: $$\text{distance} = \sqrt{(240)^2 + (40)^2} = \sqrt{57600 + 1600} = \sqrt{59200}$$ Calculate: $$\sqrt{59200} \approx 243.29$$ Distance between ball and cliff top when landed is approximately 243.29 metres. **Final answers:** (i) 40 metres (ii) 43.75 metres (iii) 45 metres (iv) 4 seconds (v) 243.29 metres