A ball which has a mass of 2 kg is dropped vertically with the initial speed of 24 m/s in the direction that forms the angle 30 degree with the horizon ,from a 40 meters tall cliff. Find a) The initial kinetic energy of the ball My solution is Ek1=(m*v^2)/2 here i replace m=2 kg and v=24 m/s and i find Ek1. b) its kinetic energy and potential energy when he is in the highest point of its trajectory. When he is at the highest point, v=0 so the Kinetic energy is zero..as for the potential energy..what am I supposed to do? c) Find the kinetic energy of the ball when he touches the ground... the kinetic energy supposed to be E=(m*v^2)/2 ,by the law of the conservation of the energy, I have that Ep(b)=Ekc+Epc where Epb is the potential energy in the point B,and Ekc and Epb are the Kinetic and potential energy in the point c d)Find the speed of the ball when it hits the ground...Here Ep=0..I have to find Ek but how?