A tennis ball is dropped on the floor from a height of 20m. It rebounds to a height of 5m. If the ball was in contact with the floor for 0.01s. What was its average acceleration during contact? `(g=10m//s^(2))`

Options

A. 300 `m//s^(2)`
B. `2000 m//s^(2)`
C. `1000 m//s^(2)`
D. `500m//s^(2)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the velocity just before the ball hits the ground.
We can use the equation of motion:
\[ v^2 = u^2 + 2gh \]
Where:
- \( u = 0 \) (initial velocity when dropped)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 20 \, \text{m} \) (height from which the ball is dropped)

Substituting the values:
\[ v^2 = 0 + 2 \times 10 \times 20 \]
\[ v^2 = 400 \]
\[ v = \sqrt{400} = 20 \, \text{m/s} \]

### Step 2: Calculate the velocity just after the ball rebounds.
Using the same equation of motion for the rebound:
\[ v'^{2} = u'^{2} - 2gh' \]
Where:
- \( u' = 0 \) (initial velocity when it starts going up)
- \( g = 10 \, \text{m/s}^2 \)
- \( h' = 5 \, \text{m} \) (height to which the ball rebounds)

Substituting the values:
\[ v'^{2} = 0 + 2 \times 10 \times 5 \]
\[ v'^{2} = 100 \]
\[ v' = \sqrt{100} = 10 \, \text{m/s} \]

### Step 3: Calculate the change in velocity (\( \Delta v \)).
The change in velocity during the contact with the floor is:
\[ \Delta v = v' - (-v) = 10 - (-20) = 10 + 20 = 30 \, \text{m/s} \]

### Step 4: Calculate the average acceleration during contact.
Average acceleration (\( a_{avg} \)) can be calculated using the formula:
\[ a_{avg} = \frac{\Delta v}{\Delta t} \]
Where:
- \( \Delta t = 0.01 \, \text{s} \)

Substituting the values:
\[ a_{avg} = \frac{30 \, \text{m/s}}{0.01 \, \text{s}} = 3000 \, \text{m/s}^2 \]

### Final Answer:
The average acceleration of the ball during contact with the floor is \( 3000 \, \text{m/s}^2 \).

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