1. **State the problem:** Fred jumps from 40,000 feet and eventually lands at 40 feet above sea level. We want to find his speed when he hits the ground. 2. **Given data:** - Initial height: $40,000$ ft - Landing height: $40$ ft - Time to reach terminal velocity: $5$ s - Terminal velocity: $176$ ft/s - Distance fallen during acceleration to terminal velocity: $480$ ft - Freefall time at terminal velocity: $3$ min $28$ s = $208$ s - Parachute deployment time: $5.5$ s (constant speed $176$ ft/s) - Deceleration after parachute fully deployed: $8$ ft/s$^2$ 3. **Calculate total distance fallen before parachute deployment:** - Distance fallen during acceleration: $480$ ft - Distance fallen during freefall at terminal velocity: $$d = v \times t = 176 \times 208 = 36608 \text{ ft}$$ - Distance fallen during parachute deployment: $$d = 176 \times 5.5 = 968 \text{ ft}$$ - Total distance before deceleration: $$480 + 36608 + 968 = 38056 \text{ ft}$$ 4. **Calculate remaining distance to ground after parachute fully deployed:** - Total height difference: $$40000 - 40 = 39960 \text{ ft}$$ - Remaining distance: $$39960 - 38056 = 1904 \text{ ft}$$ 5. **Calculate final velocity after deceleration over remaining distance:** - Use kinematic equation: $$v_f^2 = v_i^2 - 2 a d$$ where $v_i = 176$ ft/s, $a = 8$ ft/s$^2$, $d = 1904$ ft - Substitute values: $$v_f^2 = 176^2 - 2 \times 8 \times 1904 = 30976 - 30464 = 512$$ - Calculate $v_f$: $$v_f = \sqrt{512} = 22.63 \text{ ft/s}$$ 6. **Interpretation:** Fred's speed when he hits the ground is approximately $22.63$ ft/s. **Final answer:** $$\boxed{22.63 \text{ ft/s}}$$