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
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