A ball is released from the height 3 meters and after it hits the floor it reaches the height 2 meter. A) Whats the speed of the ball in the moment when it meets the ground? My answer : V^2-V0^2=2gs and here we find V=sqrt60. What is the speed of the ball in the moment it leaves the ground? V2^2-V^2=2gh2 so V2^2- sqrt60=2g*2 and here we find V2=10 c) If the ball meets the ground for 0.02 seconds whats the direction and numerical value of the acceleration? The direction is upside down and a=(10-sqrt60)/0.02 ARE THESE CORRECT?