1. The problem states a ball is thrown horizontally at a speed of 24 m/s from a cliff and hits the ground after 4.0 seconds. We need to find the height of the cliff. 2. Since the ball is thrown horizontally, its initial vertical velocity $V_i$ is $0$ m/s. 3. The acceleration due to gravity $g$ is approximately $9.8$ m/s$^2$ downwards. 4. The vertical motion of the ball is free fall under gravity, so the height $h$ of the cliff can be found using the equation for distance under constant acceleration: $$h = V_i t + \frac{1}{2} g t^2$$ Because $V_i = 0$, this simplifies to $$h = \frac{1}{2} g t^2$$ 5. Substitute $g = 9.8$ m/s$^2$ and $t = 4.0$ s: $$h = \frac{1}{2} \times 9.8 \times (4.0)^2 = 0.5 \times 9.8 \times 16 = 78.4\text{ meters}$$ 6. Therefore, the cliff is approximately 78.4 meters high.