This problem involves projectile motion, specifically finding the minimum speed required for a person to jump horizontally from a cliff to miss a ledge below.

Given:

• Width of the ledge: $L = 2.05 \, \text{m}$
• Height difference between the top of the cliff and the ledge: $h = 9.20 \, \text{m}$

The projectile motion is horizontal, and we need to find the minimum initial horizontal speed, $v_0$, such that the person travels at least $L$ meters horizontally before hitting the ledge.

Step 1: Calculate time to fall

The vertical motion is independent of the horizontal motion. Using the formula for the vertical displacement under gravity: $y = \frac{1}{2} g t^2$ where:

• $y = h = 9.20 \, \text{m}$ is the vertical displacement,
• $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity,
• $t$ is the time to fall.

Solving for $t$: $h = \frac{1}{2} g t^2$ $t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 9.20}{9.8}} = \sqrt{1.8776} \approx 1.37 \, \text{seconds}$

Step 2: Calculate minimum horizontal speed

Now, the horizontal motion is uniform, meaning the person must cover $L = 2.05 \, \text{m}$ in the time $t$ calculated above. The horizontal distance covered is: $x = v_0 t$ Solving for $v_0$: $v_0 = \frac{L}{t} = \frac{2.05}{1.37} \approx 1.50 \, \text{m/s}$

Conclusion:

The minimum speed she must have just as she leaves the top of the cliff to miss the ledge is approximately 1.50 m/s.

Would you like further details or clarifications?

Here are 5 related questions to expand on this topic:

1. How would the speed change if the ledge were farther away?
2. What if the height of the cliff were doubled—how would that affect the minimum speed?
3. How do air resistance and friction impact this type of projectile motion?
4. How would you calculate the speed if the person jumps at an angle rather than horizontally?
5. What is the maximum time the person can be in the air before hitting the ground?

Tip: Always separate the horizontal and vertical components in projectile motion problems for easier calculation!