1. **State the problem:** A baseball is thrown at a speed of 51.7 km/h, and we want to find how much it drops due to gravity over a horizontal distance of 16 feet. 2. **Convert units:** - Speed: $51.7$ km/h to m/s: $$51.7 \times \frac{1000}{3600} = 14.36\, \text{m/s}$$ - Distance: 16 feet to meters: $$16 \times 0.3048 = 4.8768\, \text{m}$$ 3. **Use kinematic equations:** The horizontal velocity $v_x$ is constant (ignoring air resistance), so time $t$ to travel 4.8768 m is: $$t = \frac{\text{distance}}{v_x} = \frac{4.8768}{14.36} = 0.34\, \text{s}$$ 4. **Calculate vertical drop:** Using the formula for vertical displacement under gravity: $$y = \frac{1}{2}gt^2$$ where $g = 9.8\, \text{m/s}^2$. 5. **Calculate:** $$y = \frac{1}{2} \times 9.8 \times (0.34)^2 = 0.567\, \text{m}$$ 6. **Convert drop to feet:** $$0.567 \times 3.28084 = 1.86\, \text{feet}$$ **Final answer:** The baseball drops approximately **1.86 feet** over 16 feet of horizontal travel due to gravity.