Question

problem # 12

• a daring 510 n swimmer dives off a cliff with a running horizontal leap, as shown in figure 3.36. what must her minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, which is 1.75 m wide and 9.00 m below the top of the cliff?

Explanation:

Step1: 计算下落时间

竖直方向：\( y = \frac{1}{2}gt^2 \)，\( t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{2×9.00}{9.8}} ≈ 1.355 \, \text{s} \)

Step2: 计算水平速度

水平方向：\( v = \frac{x}{t} = \frac{1.75}{1.355} ≈ 1.29 \, \text{m/s} \)

Answer:

1.29 m/s