When the object is launched into the air it has a kinetic energy proportional to its velocity squared. As it travels its energy is converted into gravitational potential energy as it slows down due to the effect of gravity. This gravitational potential energy is proportional to the height it reaches above the ground. Once all its kinetic energy is converted to gravitational energy, the object will have slowed to a stop in mid air. As gravity pulls the object back to the earth, the gravitational potential energy is converted back into kinetic energy and the object accelerates until it strikes the ground at the same velocity it was launched at as all the gravitational potential energy will have been converted back into kinetic energy.In this answer we ignored air resistance as directed by the question. Were we to include air resistance we would see that some energy is lost in the process to the air and the final velocity of the object as it falls to the ground again would be less than it's launch velocity as its kinetic energy would be less than its initial kinetic energy when it was launched.Kinetic energy = (1/2)mv2m = mass (kg), v = velocity (m/s)Gravitational potential energy = mghm = mass (kg), g = gravitational field strength (N/kg), h = height reached above ground (m)If you are given some of these values you can calculate the others by rearranging these equations. A slightly more challenging question would consider if the object launched from a point with a different GPE from where it landed.