Final answer:
Using the conservation of energy, the artillery shell's speed decreases as it gains height from 100 m to 200 m, resulting in a final speed of 50 m/s when it hits the target at 200 m above the ground.
Step-by-step explanation:
The question asks about the speed of an artillery shell when it hits a target 200 m above the ground, given that at 100 m high it has a speed of 100 m/s, and we are to neglect air friction. To solve this, we use the principle of conservation of energy, which states that if no external work is done on a system, the total mechanical energy remains constant.
We know that mechanical energy is the sum of potential energy and kinetic energy. As the shell rises from 100 m to 200 m, its potential energy increases, so its kinetic energy must decrease to conserve total mechanical energy. When the shell is at 200 m in height, it will have more potential energy and less kinetic energy compared to when it was at 100 m.
Therefore, the shell's speed when it hits the target will be less than 100 m/s due to the gain in potential energy. The correct answer is (a) 50 m/s.