Final answer:
The maximum height reached by the object is approximately 45.99 km. The escape speed for a rocket leaving the moon is approximately 56.711 km/s.
Step-by-step explanation:
To calculate the maximum height reached by an object projected upward from the surface of the Earth, we can use the equations of motion. The initial speed of the object is given as 3 km/s and the acceleration due to gravity is 9.81 m/s². We can use the kinematic equation, vf² = vi² + 2ad, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and d is the displacement. At the maximum height, the object's final velocity is 0. Therefore, we can rearrange the equation to solve for the displacement, which represents the maximum height. Substituting the given values, we have:
0 = (3 km/s)² - 2(9.81 m/s²)d
Simplifying, we get:
d = (3 km/s)² / (2(9.81 m/s²))
Converting the speed to m/s, we have:
d = (3 km/s * 10³ m/km)² / (2(9.81 m/s²))
d ≈ 45986.87 m
Converting the height to km, we have:
d ≈ 45.99 km
Therefore, the maximum height reached by the object is approximately 45.99 km.
To find the escape speed for a rocket leaving the moon, we can use a similar approach. The escape speed is the minimum speed required for an object to escape the gravitational pull of an astronomical body. The escape speed from Earth is given as 11.2 km/s. On the moon, the acceleration due to gravity is 0.166 times that on Earth, and the moon's radius is 0.273 times the Earth's radius. We can use the equation for escape velocity, v = sqrt(2gr), where v is the escape velocity, g is the acceleration due to gravity, and r is the radius of the astronomical body.
Substituting the given values, we have:
v = sqrt(2(0.166 * 9.81 m/s²)(0.273 * 6.37×10⁶ m))
Simplifying, we get:
v = sqrt(3.21074×10⁶ m²/s²)
Converting the velocity to km/s, we have:
v ≈ sqrt(3.21074×10⁶ m²/s²) / 1000 m/km
v ≈ 56.711 km/s
Therefore, the escape speed for a rocket leaving the moon is approximately 56.711 km/s.