1. **State the problem:** We are given an object dropped from rest (initial speed $u=0$ m/s) falling down a distance $s=43$ m in time $t=4.4$ s with final speed $v$ unknown. 2. **Formula used:** The formula for distance traveled under constant acceleration when initial and final speeds are known is: $$s = \frac{1}{2} (u + v) t$$ where - $s$ is distance traveled, - $u$ is initial speed, - $v$ is final speed, - $t$ is time. 3. **Find the final speed $v$:** Rearrange the formula to solve for $v$: $$s = \frac{1}{2} (u + v) t \implies 2s = (u + v) t \implies \frac{2s}{t} = u + v \implies v = \frac{2s}{t} - u$$ 4. **Substitute known values:** $$v = \frac{2 \times 43}{4.4} - 0 = \frac{86}{4.4} = 19.5454545...$$ 5. **Final answer:** The final speed $v$ of the object after falling 43 m in 4.4 s is approximately: $$v \approx 19.55\ \text{m/s}$$ This means the object is moving downward at about 19.55 meters per second after 4.4 seconds of free fall.