You throw a ball downward from a window at a speed of 2.7 m/s. How fast will it be moving when it hits the sidewalk 2.9 m below?
s = 2.9 - 2.7t - 4.9t^2
s=0 when t=0.54s
v = -2.7 - 9.8t
v(.54) = -8m/s
To find the speed at which the ball will hit the sidewalk, we can use the equations of motion.
Let's denote:
- Initial velocity (upward) as u = +2.7 m/s
- Final velocity (downward) as v = ?
- Acceleration due to gravity as g = -9.8 m/s² (negative sign because it acts downward)
- Displacement as s = -2.9 m (negative sign because it is downward)
The equation we can use to solve for the final velocity is:
v² = u² + 2as
First, let's plug in the known values:
v² = (2.7 m/s)² + 2(-9.8 m/s²)(-2.9 m)
Simplifying:
v² = 7.29 m²/s² + 56.84 m²/s²
v² = 64.13 m²/s²
To find v, let's take the square root of both sides of the equation:
v = √(64.13 m²/s²)
After calculating, we find that v ≈ 8.01 m/s.
Therefore, when the ball hits the sidewalk 2.9 m below the window, it will be moving at a speed of approximately 8.01 m/s.