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/s². (Given g =10 m/s²)

Hint:

For average acceleration:

• Calculate the velocities before impact and after rebound using \(v = \sqrt{2gh}\).
• Use \(\Delta v = v_\text{before} + v_\text{after}\) for the total change in velocity.

Answer 1

Correct Answer: 120

1. Velocity Just Before Impact: Using \(v^2 = u^2 + 2gh\), where \(u = 0\), \(h = 9.8 \, \text{m}\):

   \[ v = \sqrt{2 \cdot 10 \cdot 9.8} = \sqrt{196} = 14 \, \text{m/s}. \]

2. Velocity Just After Rebound: Using \(v^2 = u^2 + 2gh\), where \(u = 0\), \(h = 5.0 \, \text{m}\):

   \[ v = \sqrt{2 \cdot 10 \cdot 5.0} = \sqrt{100} = 10 \, \text{m/s}. \]

3. Change in Velocity During Contact: Total change in velocity:

   \[ \Delta v = v_\text{before impact} + v_\text{after rebound} = 14 + 10 = 24 \, \text{m/s}. \]

4. Average Acceleration: Average acceleration:

   \[ a = \frac{\Delta v}{\Delta t} = \frac{24}{0.2} = 120 \, \text{m/s}^2. \]

Final Answer: 120 m/s²