You throw a ball upward from a window at a speed of 2.9 m/s. How fast will it be moving when it hits the sidewalk 2.2 m below?

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A person leans over the side of a deep well and drops a stone that falls freely. After 1.0s she hears the splash of the stone striking the water at the bottom of the well. How deep is the well?

To find the speed at which the ball will be moving when it hits the sidewalk, we can use the equations of motion.

First, let's list the known values:
- Initial velocity (u) = 2.9 m/s (upward)
- Final velocity (v) = ?
- Acceleration (a) = -9.8 m/s² (acceleration due to gravity, directed downward)
- Displacement (s) = -2.2 m (downward)

We can use the equation: v² = u² + 2as

Substituting the known values:
v² = (2.9 m/s)² + 2(-9.8 m/s²)(-2.2 m)

Simplifying the equation:
v² = 8.41 m²/s² + 42.96 m²/s²
v² = 51.37 m²/s²

Now, let's take the square root of both sides to find v:
v = √(51.37 m²/s²)
v ≈ 7.17 m/s

Therefore, when the ball hits the sidewalk, it will be moving at approximately 7.17 m/s.