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 ___________ ms⁻².
(Given g = 10 ms⁻²)

Solution

Physics sir · 4 steps

Velocity flips direction at the floor, so the change is the sum of the two speeds, not their difference.

1. Speed on impact. The ball falls freely through 9.8 m, so v₁ = √2gh₁ = √(2 × 10 × 9.8) = √196 = 14 m s⁻¹, directed downward.
2. Speed on rebound. To rise 5 m, it must leave the floor with v₂ = √2gh₂ = √(2 × 10 × 5) = 10 m s⁻¹, directed upward.
3. Change in velocity. Take upward as positive. Then v₁ = -14 m s⁻¹ and v₂ = +10 m s⁻¹, so Δ v = 10 - (-14) = 24 m s⁻¹.

   a_(avg) = (Δ v)/(Δ t) = (24 m s⁻¹)/(0.2 s) = 120 m s⁻²

4. So. The average acceleration during contact is 120 m s⁻², directed upward.

The common slip: 20. Subtracting the speeds as 14 - 10 = 4 m s⁻¹ ignores the reversal in direction and gives 20, not 120.

The idea to keep. Acceleration is the change in velocity, and velocity is a vector. When a ball bounces, it reverses direction, so you add the impact and rebound speeds.

Answer from the NTA final answer key. Solution by PrepWiser.