A daring 510 N swimmer dives off a cliff with a running horizontal leap. 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?

To solve this problem, we will break it down into two components: horizontal and vertical motion, treating the swimmer's leap as a projectile motion problem.

### Given:
- Weight of the swimmer \( W = 510 \, \text{N} \)
- Width of the ledge (horizontal distance to clear) \( d_x = 1.75 \, \text{m} \)
- Height difference between the top and bottom of the cliff (vertical distance) \( d_y = 9.00 \, \text{m} \)

The minimum speed required is the swimmer's **horizontal velocity** as she leaves the cliff. The key concept here is that the horizontal motion is uniform (constant speed), while the vertical motion is accelerated due to gravity.

### Step 1: Time to fall
We first calculate the time \( t \) it takes for the swimmer to fall the vertical distance \( d_y \) of 9.00 meters under the influence of gravity.

Using the kinematic equation for vertical motion:
\[
d_y = \frac{1}{2} g t^2
\]
where:
- \( d_y = 9.00 \, \text{m} \) (vertical distance)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

Solving for \( t \):
\[
t = \sqrt{\frac{2 d_y}{g}} = \sqrt{\frac{2 \times 9.00}{9.8}} = \sqrt{1.8367} \approx 1.355 \, \text{seconds}
\]

### Step 2: Minimum horizontal speed
Now, we calculate the horizontal speed required to cover the horizontal distance \( d_x \) of 1.75 meters in the time \( t \) found above.

Since horizontal velocity \( v_x \) is constant, we use the equation:
\[
d_x = v_x t
\]
Solving for \( v_x \):
\[
v_x = \frac{d_x}{t} = \frac{1.75 \, \text{m}}{1.355 \, \text{s}} \approx 1.29 \, \text{m/s}
\]

### Final Answer:
The swimmer must have a minimum horizontal speed of approximately **1.29 m/s** to safely clear the ledge.

Would you like further details or have any questions? Here are five related questions to explore this further:

1. How would the required horizontal speed change if the height of the cliff increased?
2. What happens if the swimmer has a higher horizontal speed than the minimum?
3. How does air resistance affect the motion of the swimmer?
4. Can we determine the initial velocity of the swimmer if she jumps at an angle?
5. How would the horizontal speed requirement change if the ledge were wider?

**Tip:** Always separate the horizontal and vertical components when solving projectile motion problems; they are independent of each other!

Find out how to calculate the minimum speed a swimmer needs to clear a ledge 1.75 meters wide and 9.00 meters below the top of a cliff. This problem uses kinematic equations to solve for horizontal and vertical motion, making it a great example of projectile motion for students in Grades 9-12.

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