A 222 kg weather rocket reaches its maximum height 32 km above the Earth's surface. Determine:
A.) The amount of gravitational potential energy at its maximum height.
B.)Its speed when it hits the ground (neglect frictional effects).
To determine the amount of gravitational potential energy at its maximum height, we can use the formula:
Gravitational Potential Energy = mass * gravity * height
Where:
mass = 222 kg
gravity = 9.8 m/s^2 (acceleration due to gravity)
height = 32 km = 32,000 m
Gravitational Potential Energy = 222 kg * 9.8 m/s^2 * 32,000 m
Gravitational Potential Energy = 68,710,400 J
So, the amount of gravitational potential energy at its maximum height is 68,710,400 J.
To determine its speed when it hits the ground, we can use the principle of conservation of energy. At maximum height, the gravitational potential energy is converted completely to kinetic energy:
Kinetic Energy = Gravitational Potential Energy
Therefore:
1/2 * mass * velocity^2 = Gravitational Potential Energy
Solving for velocity:
velocity^2 = (2 * Gravitational Potential Energy) / mass
velocity^2 = (2 * 68,710,400 J) / 222 kg
velocity^2 ≈ 620,180 m^2/s^2
Taking the square root of both sides:
velocity ≈ √(620,180 m^2/s^2)
velocity ≈ 787.4 m/s
So, the speed of the rocket when it hits the ground (neglecting frictional effects) is approximately 787.4 m/s.