A ball dropped on to the floor from a height of 10 m rebounds to a height of 2.5 m. If the ball is in contact with the floor for 0.02s, its average acceleration during contact is

Given:

• Initial height (h₁): 10 m
• Rebound height (h₂): 2.5 m
• Contact time (Δt): 0.02 s

Step 1: Calculate the velocity just before impact (v₁) and just after rebound (v₂).

Using the kinematic equation: v = sqrt(2gh), where:

• g: acceleration due to gravity (approximately 9.8 m/s²).
• v₁ (velocity just before impact):v₁ = sqrt(2 * 9.8 * 10) = 14.0 m/s
• v₂ (velocity just after rebound):v₂ = sqrt(2 * 9.8 * 2.5) = 7.0 m/s

Step 2: Calculate the change in velocity (Δv):

Δv = v₂ - (-v₁) = v₂ + v₁

Step 3: Calculate the average acceleration (a) during contact:

a = Δv / Δt

Calculation:

• Δv = 7.0 m/s + 14.0 m/s = 21.0 m/s
• a = 21.0 m/s / 0.02 s = 1050 m/s²

Final Answer

The average acceleration during contact is 1050 m/s².