Final answer:
The kinetic energy of an object just before it hits the ground will be equal to the potential energy it had initially if air resistance is ignored. Therefore, if the object initially had a potential energy of 387 J, it will have a kinetic energy of 387 J just before impact.
Step-by-step explanation:
The question is asking about the relationship between potential energy (PE) and kinetic energy (KE) of an object that is dropped from a height just before it hits the ground. According to the principle of conservation of mechanical energy, in the absence of air resistance, the potential energy that the object has at the beginning of its fall will be converted into kinetic energy just before it reaches the ground.
In mathematical terms, if we ignore air resistance, the potential energy (PE) of an object at a certain height is given by PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height from which the object is dropped. As the object falls, this potential energy is converted into kinetic energy (KE), which can be represented as KE = ½mv², where m is the mass of the object and v is its velocity.
Since PE = KE when ignoring air resistance, the kinetic energy of the object just before hitting the ground will be equal to the potential energy it had at the beginning of the fall. Therefore, if an object has a potential energy of 387 J at a certain height, it will have a kinetic energy of 387 J in the instant before it hits the ground.