Final answer:
The height above the Earth that the Space Shuttle would be given the escape velocity V = 2.45 km/s, the gravitational pull g = 0.0098 km/sec, and the radius of the Earth r = 6440 km is approximately 1622.857 kilometers.
Step-by-step explanation:
The height above the Earth that the Space Shuttle would be given the escape velocity V = 2.45 kilometers/second, the gravitational pull g = 0.0098 km/sec, and the radius of the Earth r = 6440 km can be calculated using the formula:
h = (r^2 * g) / (2 * V^2)
Substituting the given values into the formula, h = (6440^2 * 0.0098) / (2 * 2.45^2) = 1622.857 km.
Therefore, the height above the Earth that the Space Shuttle would be is approximately 1622.857 kilometers.