A soccer ball accelerates from rest and rolls 6.5 meters down a hill in 3.1 seconds. It then bumps a tree. What is the speed of the ball right before it hits a tree?

To find the speed of the ball right before it hits the tree, we can use the concept of constant acceleration.

First, we need to find the acceleration of the ball down the hill. We can use the equation of motion:
\[s = ut + \frac{1}{2}at^2\]
where:
- \(s\) is the distance traveled (6.5 meters),
- \(u\) is the initial velocity (rest, so \(u = 0\)),
- \(t\) is the time taken (3.1 seconds), and
- \(a\) is the acceleration that needs to be determined.

Since the ball rolls down the hill, we know that gravity is acting as the acceleration. The acceleration due to gravity on the Earth is approximately \(9.8 \, \text{m/s}^2\). So we can replace \(a\) with \(9.8 \, \text{m/s}^2\). Substituting the values into the equation, we get:
\[6.5 = 0 + \frac{1}{2} \cdot 9.8 \cdot (3.1)^2\]

Simplifying the equation, we have:
\[6.5 = 0 + 4.9 \cdot (3.1)^2\]

To find the speed of the ball before it hits the tree, we need to know the final velocity. Using the equation of motion:
\[v = u + at\]
we rearrange the equation to solve for \(v\):
\[v = at\]

Since the ball has accelerated for the entire 3.1 seconds down the hill, we can use the same acceleration due to gravity, \(a = 9.8 \, \text{m/s}^2\), and the time, \(t = 3.1 \, \text{seconds}\), to calculate the velocity. Substituting the values:
\[v = (9.8 \, \text{m/s}^2) \cdot (3.1 \, \text{seconds})\]
\[v = 30.38 \, \text{m/s}\]

Therefore, the speed of the soccer ball right before it hits the tree is approximately 30.38 m/s.