Question Detail
Question
Two balls with equal mass are dropped from a height of 2 meters above the floor. They both bounce back up to a maximum height of 1.8 meters above the floor. The first ball is made of steel and is in contact with the floor for 0.04 seconds. The second ball is made of rubber and is in contact with the floor for 0.09 seconds. What is the speed of the steel ball in m/s just before it hits the floor? (acceleration due to gravity, g = 9.81 m/s²)
Correct Answer
Option C is the correct answer.
Detailed Explanation
The speed just before hitting the floor is found from v² = u² + 2gh, with u=0, h=2 m, g=9.81. So v = √(2×9.81×2) = √(39.24) = 6.264 m/s ≈ 6.26 m/s.
Hint
Use conservation of energy or kinematics to find velocity just before impact.