A tennis ball is dropped on to the floor from a height of 9.8 m. It rebounds to a height 5.0 m. Ball comes in contact with the floor for 0.2 s. The average acceleration during contact is ____ m s⁻². [Given g = 10 m s⁻²]
Physics · Circular Motion
JEE Mains 29 January 2023 Shift 1 Morning Shift

Question

A tennis ball is dropped on to the floor from a height of 9.8 m. It rebounds to a height 5.0 m. Ball comes in contact with the floor for 0.2 s. The average acceleration during contact is ____ m s⁻². [Given g = 10 m s⁻²]

Physics · Circular Motion

JEE Mains 29 January 2023 Shift 1 Morning Shift

G1 Physics · Accepted Answer

Speed just before impact = √(2gh₁) = √(2×10×9.8) = 14 m s⁻¹ downward. Rebound speed = √(2gh₂) = √(2×10×5) = 10 m s⁻¹ upward. Change in velocity magnitude = 24 m s⁻¹. Therefore average acceleration magnitude = 24/0.2 = 120 m s⁻².