The image shows a physics problem involving a ball rolling off the edge of a horizontal platform. Here's a breakdown of the given information:

• A ball is rolling horizontally with a speed of $v = 15 \, \text{m/s}$.
• The ball falls off the edge of a platform that is $6.0 \, \text{m}$ high.
• It lands on another platform that is $2.6 \, \text{m}$ high.
• The question asks for the horizontal distance the ball travels while falling from the 6.0 m platform to the 2.6 m platform.

Solution:

1. Determine the time the ball takes to fall vertically:

   The vertical distance the ball falls is $h = 6.0 - 2.6 = 3.4 \, \text{m}$.

   The equation for the vertical displacement under gravity is: $h = \frac{1}{2} g t^2$

   Solving for $t$ (time of fall): $t = \sqrt{\frac{2h}{g}}$

   Plugging in the values: $t = \sqrt{\frac{2 \times 3.4}{9.8}}$

2. Calculate the horizontal distance traveled:

   The horizontal distance $d$ the ball travels is given by: $d = v \cdot t$ where $v = 15 \, \text{m/s}$ is the horizontal speed of the ball.

Now, let's calculate the values.The time it takes for the ball to fall is approximately $t \approx 0.83 \, \text{seconds}$.

The horizontal distance the ball travels before landing on the lower platform is approximately $d \approx 12.5 \, \text{meters}$.

Would you like more details, or do you have any questions?

Here are some related questions for further exploration:

1. What would happen if the initial speed of the ball were doubled?
2. How would the time of fall change if the height difference was increased to 5 meters?
3. What is the impact of air resistance on the horizontal distance traveled?
4. How would you calculate the horizontal distance if the ball were launched at an angle?
5. Can you determine the velocity of the ball just before it lands on the lower platform?

Tip: Understanding the components of projectile motion can help solve problems where objects are in free fall combined with horizontal motion.