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 0.2s. The average acceleration during contact is __________ .
(Given )
(Given )
Show answer and solution
Answer: 120
Solution
Just before hitting floor: v₁ = √(2×10×9.8) = 14 m/s (downward)
Just after rebound: v₂ = √(2×10×5.0) = 10 m/s (upward)
Taking upward as positive, change in velocity = v₂ − (−v₁) = 10 + 14 = 24 m/s
Average acceleration = Δv/Δt = 24/0.2 = 120 ms⁻² (directed upward).
Just after rebound: v₂ = √(2×10×5.0) = 10 m/s (upward)
Taking upward as positive, change in velocity = v₂ − (−v₁) = 10 + 14 = 24 m/s
Average acceleration = Δv/Δt = 24/0.2 = 120 ms⁻² (directed upward).
[JEE Main 2023 · 29 Jan · Shift 1]