A rigid body of mass 2 kg is dropped from a stationary balloon kept at a height of 50 m from the ground. The speed of the body when it just touches the ground and the total energy when it is dropped from the balloon are respectively (acceleration due to gravity = 9.8 m/s²)
980 m.s⁻¹ and 980 J
√980 m.s⁻¹ and √980 J
980 m.s⁻¹ and √980 J
√980 m.s⁻¹ and 980 J
Answer is Wrong!
Answer is Right!
This question was previously asked in
UPSC NDA-2 – 2019
1. Speed when it just touches the ground (v): Using the kinematic equation v² = u² + 2gh, we get v² = 0² + 2 * 9.8 * 50 = 980. So, v = √980 m/s.
2. Total energy when dropped: At the moment of dropping from a height h with zero initial velocity, the total mechanical energy is the sum of potential energy and kinetic energy. Potential Energy (PE) = mgh = 2 kg * 9.8 m/s² * 50 m = 980 J. Kinetic Energy (KE) = ½mu² = ½ * 2 kg * (0 m/s)² = 0 J. Total Energy = PE + KE = 980 J + 0 J = 980 J. (Note: By conservation of energy, the total energy remains 980 J just before hitting the ground, where all potential energy is converted to kinetic energy).