since it already reaches the ground, it should come to a stop. therefore, v should be 0, and KE=0. right?If an object reaches the ground, how much is the velocity and kinetic energy?
Once the object stops, then yes...PE = KE = o.If an object reaches the ground, how much is the velocity and kinetic energy?
You are right, but... this is important... problems like this usually mean what is the velocity and KE just prior to hitting the ground. A fraction of a fraction of a centimeter above the ground for example.
In this case, without air drag or other forces, the potential energy from the release height H is PE = mgH = 1/2 mV^2 = KE just prior to impact. That is, all the PE is converted into KE just prior to hitting the ground. This comes from the conservation of energy law, which is what this kind of problems is addressing.
Note, if you cross out the mass m in the above equation, one gets 2gH = V^2 which is a variation of one of the SUVAT motion equations. And that's where it comes from... the conservation of energy law.
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