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)
- A.980 m s⁻¹ and 980 J
- B.√980 m s⁻¹ and √980 J
- C.980 m s⁻¹ and √980 J
- D.√980 m s⁻¹ and 980 J
Correct answer
Explanation
The correct answer is D, √980 m s⁻¹ and 980 J. The body falls freely from rest through 50 m, so v² = 2gh = 2 × 9.8 × 50 = 980, giving v = √980, about 31.3 m s⁻¹; its total energy at release is all potential energy, mgh = 2 × 9.8 × 50 = 980 J. Because the balloon is stationary, the body starts with zero velocity and zero kinetic energy. As it falls, potential energy turns into kinetic energy, but the total stays 980 J if air resistance is ignored. At the ground, ½mv² = ½ × 2 × 980 = 980 J. Option A is wrong because the speed is the square root of 980, not 980 itself; 980 m s⁻¹ would be nearly three times the speed of sound. Option B is wrong because energy is mgh, which is 980 J, not its square root. Option C is wrong on both counts: the speed should carry the square root and the energy should not. Exam tip: free fall from rest gives v = √(2gh); total mechanical energy = mgh at the top = ½mv² at the bottom.
Practice Questions
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Show answer
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