1. Problem 22: A ball is thrown vertically upward with an initial speed of 25 m/s. We need to find the time to reach the highest point. 2. Use the formula for velocity under constant acceleration: $$v_f = v_i + gt$$ where $v_f$ is final velocity, $v_i$ is initial velocity, $g$ is acceleration due to gravity ($-10$ m/s², negative because it acts downward), and $t$ is time. 3. At the highest point, the velocity $v_f = 0$. Substitute values: $$0 = 25 + (-10) t$$ 4. Solve for $t$: $$-25 = -10 t \implies t = \frac{25}{10} = 2.5 \text{ seconds}$$ 5. Answer for problem 22 is 2.5 seconds (Option A). --- 6. Problem 23: A 3.0 kg mass is thrown upward with initial speed 10 m/s. Find the maximum height reached. 7. Use the kinematic formula: $$v_f^2 = v_i^2 + 2 g h$$ At the highest point, $v_f = 0$, so: $$0 = (10)^2 + 2 (-10) h$$ 8. Solve for $h$: $$0 = 100 - 20 h \implies 20 h = 100 \implies h = 5 \text{ meters}$$ 9. The mass reaches a height of 5 meters. --- 10. Problem 24: A rock is dropped from a cliff and hits the ground with velocity 80 m/s. Find the height of the cliff. 11. Use the formula: $$v_f^2 = v_i^2 + 2 g h$$ Since the rock is dropped, $v_i = 0$, $v_f = 80$ m/s, $g = 10$ m/s² (downward), solve for $h$: 12. Substitute: $$80^2 = 0 + 2 \times 10 \times h \implies 6400 = 20 h \implies h = 320 \text{ meters}$$ 13. The height of the cliff is 320 meters, which is not among the options given (0.05, 10, 20, 40 m). Possibly a typo or different gravity assumed. --- 14. Problem 25: A person drops a pebble down a well and hears the sound 3 seconds later. The well is 5.0 m high above the pebble to reach? The problem statement is incomplete, so cannot solve. Final answers: - Problem 22: 2.5 seconds - Problem 23: 5 meters - Problem 24: 320 meters (not matching options) - Problem 25: Insufficient data