A kid slides down a 1.7 m tall water slide into a pool, where the water level is 29 cm below the bottom of the slide. At what horizontal position from the edge of the slide does the kid land in the water?

This problem can be solved using basic kinematics. We'll approach it step-by-step:

### Step 1: Determine the vertical distance the kid falls
The total vertical distance the kid falls is the height of the slide plus the difference between the bottom of the slide and the water level. 

\[
\text{Total vertical distance} = 1.7 \, \text{m} + 0.29 \, \text{m} = 1.99 \, \text{m}
\]

### Step 2: Time to fall
The kid falls freely under gravity, so we can use the kinematic equation for vertical motion to find the time it takes to fall:

\[
y = \frac{1}{2} g t^2
\]

Where:
- \( y = 1.99 \, \text{m} \) is the total vertical distance,
- \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity,
- \( t \) is the time it takes to fall.

Solving for \( t \):

\[
t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{2 \times 1.99}{9.8}} = \sqrt{0.406} \approx 0.64 \, \text{seconds}
\]

### Step 3: Horizontal distance traveled
Assuming the kid leaves the slide horizontally with an initial horizontal velocity \( v_x \), the horizontal distance they travel is given by:

\[
x = v_x t
\]

Since the time \( t \) has already been calculated, we just need to find the horizontal velocity \( v_x \).

### Step 4: Finding horizontal velocity
If we assume the kid slides down the entire slide with no friction, we can use energy conservation to find \( v_x \). The potential energy at the top of the slide is converted into kinetic energy at the bottom.

\[
mgh = \frac{1}{2} m v_x^2
\]

Where:
- \( h = 1.7 \, \text{m} \) (height of the slide),
- \( g = 9.8 \, \text{m/s}^2 \),
- \( v_x \) is the horizontal velocity.

Solving for \( v_x \):

\[
v_x = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 1.7} = \sqrt{33.32} \approx 5.77 \, \text{m/s}
\]

### Step 5: Calculate horizontal position
Now, we can find the horizontal distance \( x \) by plugging in \( v_x \) and \( t \):

\[
x = v_x \times t = 5.77 \times 0.64 \approx 3.69 \, \text{m}
\]

### Final Answer:
The kid will land approximately **3.69 meters** horizontally from the edge of the slide.

Would you like further details or clarifications on any of the steps? 

Here are 5 related questions you might find interesting:
1. How would the landing distance change if the height of the slide increased?
2. What if the water level was the same as the bottom of the slide—how would this affect the landing position?
3. How does air resistance affect the trajectory of the kid?
4. Could the kid's mass alter the result in any way?
5. How can we determine the velocity at which the kid hits the water?

**Tip:** In free-fall problems, the horizontal velocity remains constant if there's no air resistance!

Solve how far a kid lands from the edge of a 1.7 m water slide using kinematics and energy conservation. The problem involves free fall, projectile motion, and basic physics formulas, making it ideal for Grades 9-12 students.

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