You throw a ball downward from a window at a speed of 1.9 m/s. How fast will it be moving when it hits the sidewalk 2.2 m below?
final KE is equal to the initial KE added to the initial PE
1/2 m vf^2=1/2 m vi^2 + mg (2.2)
solve for vf
i don't understand what mg is..we havent learned that?
jjnj
To find out how fast the ball will be moving when it hits the sidewalk, we can use the equations of motion.
First, let's denote the initial velocity of the ball as "v0," which in this case is 1.9 m/s (downward velocity). The acceleration due to gravity is denoted as "g" and is approximately 9.8 m/s² (assuming negligible air resistance). The distance traveled by the ball, denoted as "d," is 2.2 m (downward distance). And we need to find the final velocity of the ball, which can be denoted as "vf."
Since the acceleration due to gravity is acting downward, we can consider it as a negative value. So, the value of "g" will be -9.8 m/s² in this case.
We can use the following equation of motion to solve for the final velocity:
vf² = v0² + 2*g*d
Substituting the given values:
vf² = (1.9 m/s)² + 2 * (-9.8 m/s²) * (2.2 m)
vf² = 3.61 m²/s² - 43.12 m²/s²
vf² = -39.51 m²/s²
As we calculated, the square of the final velocity is -39.51 m²/s². However, since velocity cannot be negative in this context (it represents the magnitude of the velocity), we take the square root of both sides to find the positive value of the final velocity:
vf = √(-39.51 m²/s²)
vf ≈ 6.28 m/s
Therefore, the ball will be moving at approximately 6.28 m/s just before it hits the sidewalk.