1. **State the problem:** Find the maximum height reached by an object projected upwards with initial velocity $u=22$ m/s under acceleration $a=-10$ m/s², and then find its height 1.8 seconds after it starts descending. 2. **Find maximum height using equation $v^2 = u^2 + 2as$:** Given: $v=0$ (velocity at max height), $u=22$, $a=-10$ Substitute into formula: $$0^2 = 22^2 + 2 \times (-10) \times s$$ Simplify: $$0 = 484 - 20s$$ Rearrange: $$20s = 484$$ $$s = \frac{484}{20} = 24.2 \text{ meters}$$ This is the maximum height. 3. **Find height 1.8 s later on the way down:** After reaching max height, initial velocity $u=0$ (momentarily at rest), acceleration $a=10$ m/s² (downwards), time $t=1.8$ s. Use formula: $$s = ut + \frac{1}{2} a t^2 = 0 + \frac{1}{2} \times 10 \times (1.8)^2$$ Calculate: $$s = 5 \times 3.24 = 16.2 \text{ meters}$$ This is the distance fallen from max height. 4. **Calculate height above ground:** $$\text{Height} = \text{max height} - s = 24.2 - 16.2 = 8 \text{ meters}$$ **Explanation about dividing time by 2:** The time was divided by 2 when calculating velocity at the peak, to find that peak point since the total time of flight splits evenly into ascent and descent. But here, $t=1.8$ s is a time after the peak during descent; it is not divided. We use the time after peak directly in the formula for displacement with initial velocity zero because the motion starts anew from rest at max height going downward. Final answers: - Maximum height $= 24.2$ m - Height 1.8 seconds after max height $= 8$ m