1. **Problem Statement:** Bob throws a ball from the top of a cliff with initial speed $v_0 = 32.1$ m/s. The ball returns to the same height after $t_1 = 0.710$ s. The ball lands on the beach at horizontal distance $x = 129$ m from the cliff base. Find the height $h$ of the cliff. 2. **Relevant formulas and concepts:** - Vertical motion under gravity: $y(t) = h + v_{0y} t - \frac{1}{2} g t^2$ - Horizontal motion: $x = v_{0x} t_{total}$ - Gravity acceleration: $g = 9.8$ m/s$^2$ - At time $t_1$, ball returns to initial height: $y(t_1) = h$ 3. **Step 1: Find vertical component of initial velocity $v_{0y}$** - Since ball returns to initial height at $t_1$, vertical displacement is zero: $$0 = v_{0y} t_1 - \frac{1}{2} g t_1^2$$ - Solve for $v_{0y}$: $$v_{0y} = \frac{1}{2} g t_1 = \frac{1}{2} \times 9.8 \times 0.710 = 3.479 \text{ m/s}$$ 4. **Step 2: Find total time of flight $t_{total}$** - The ball lands on the ground at horizontal distance $x = 129$ m. - Horizontal velocity component $v_{0x}$ is unknown, but horizontal velocity is constant: $$v_{0x} = \frac{x}{t_{total}}$$ 5. **Step 3: Use Pythagoras to find $v_0$ relation:** - Initial speed magnitude: $$v_0^2 = v_{0x}^2 + v_{0y}^2$$ - Rearranged: $$v_{0x} = \sqrt{v_0^2 - v_{0y}^2} = \sqrt{32.1^2 - 3.479^2} = \sqrt{1030.41 - 12.10} = \sqrt{1018.31} = 31.91 \text{ m/s}$$ 6. **Step 4: Calculate total time of flight $t_{total}$** $$t_{total} = \frac{x}{v_{0x}} = \frac{129}{31.91} = 4.04 \text{ s}$$ 7. **Step 5: Use vertical motion to find height $h$** - Vertical position at landing time $t_{total}$ is zero (ground level): $$0 = h + v_{0y} t_{total} - \frac{1}{2} g t_{total}^2$$ - Solve for $h$: $$h = \frac{1}{2} g t_{total}^2 - v_{0y} t_{total} = \frac{1}{2} \times 9.8 \times (4.04)^2 - 3.479 \times 4.04$$ $$= 0.5 \times 9.8 \times 16.32 - 14.06 = 79.94 - 14.06 = 65.88 \text{ m}$$ **Final answer:** $$\boxed{h = 65.9 \text{ meters}}$$