A ball is dropped from rest at a height of 2 metres. Assuming acceleration due to gravity (g) is 10m/s^2 calculate the velocity of the ball just before it hits the floor.

To calculate the velocity of the ball just before it hits the floor, we can use the kinematic equation for motion under constant acceleration. Since the ball is dropped from rest, its initial velocity (u) is 0 m/s. The equation we will use is: v^2 = u^2 + 2gh Here, v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and h is the height from which the ball is dropped. Substituting the given values: u = 0 m/s (since the ball is dropped from rest), g = 10 m/s^2 (acceleration due to gravity), h = 2 m (height), The equation becomes: v^2 = 0^2 + 2 * 10 m/s^2 * 2 m = 40 m^2/s^2 To find v, take the square root of both sides: v = sqrt(40) m/s ≈ 6.32 m/s So, the velocity of the ball just before it hits the floor is approximately 6.32 m/s.