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.

Using the equation V² = U² + 2as, where V = final velocity, U = initial velocity, a = acceleration, s = displacement, we can substitute in the values given in the question to calculate the final velocity.U = 0 m/s as dropped from resta = 10m/s²s = 2 metresRearranging to solve for V gives:V = sqrt (U² + 2as) Substituting in values:V = sqrt( 0² + 2102) V = 6.32m/s