Final answer:
The theoretical landing speed of the space shuttle glider in the absence of air resistance can be determined by applying conservation of mechanical energy. The glider would convert its entire potential energy at launch into kinetic energy at landing, resulting in a higher landing speed than the 200 km/h observed when considering air resistance.
Step-by-step explanation:
To determine the landing speed of the space shuttle glider in the absence of air resistance, we can use principles from physics related to energy conservation. When air resistance is negligible, mechanical energy is conserved. The initial mechanical energy of the glider, which is the sum of its potential and kinetic energies at launch, would equal its mechanical energy at landing.
At a height of 3300 meters, the potential energy given by PE = mgh, where 'm' is mass, 'g' is the acceleration due to gravity (9.80 m/s²), and 'h' is the height, would be transformed into kinetic energy as the glider descends, given by KE = 0.5mv², where 'v' is the velocity. Therefore, upon landing, the glider would have converted all its initial potential energy into kinetic energy, resulting in a higher speed than the observed 200 km/h if there were no air resistance to dissipate energy.
The kinetic energy at landing would equal the sum of the potential energy the glider had at launch and its initial kinetic energy minus any energy lost due to air resistance, which in the absence of air resistance is zero. Using this approach, we could calculate the theoretical landing speed of the glider in a frictionless environment. It would be higher than the 200 km/h actual landing speed, which was slowed down due to the resistance of the air during the descent.