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 of the average acceleration of the tennis ball during its contact with the floor, we will follow these steps:

### Step 1: Determine the velocity just before impact (v1)
The ball is dropped from a height of 20 meters. We can use the equation of motion to find the velocity just before it strikes the ground. The relevant equation is:

\[ v^2 = u^2 + 2gh \]

Where:
- \( v \) = final velocity (just before impact)
- \( u \) = initial velocity (0 m/s, since it is dropped)
- \( g \) = acceleration due to gravity (10 m/s²)
- \( h \) = height (20 m)

Substituting the values:

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

So, the velocity just before impact (v1) is 20 m/s downward.

### Step 2: Determine the velocity just after rebound (v2)
The ball rebounds to a height of 5 meters. We can again use the equation of motion to find the velocity just after it leaves the ground. The relevant equation is:

\[ v^2 = u^2 - 2gh \]

Where:
- \( v \) = final velocity (0 m/s at the peak of the rebound)
- \( u \) = initial velocity (just after rebound)
- \( g \) = acceleration due to gravity (10 m/s²)
- \( h \) = height (5 m)

Rearranging gives:

\[ 0 = u^2 - 2 \times 10 \times 5 \]
\[ u^2 = 100 \]
\[ u = \sqrt{100} = 10 \, \text{m/s} \]

So, the velocity just after rebound (v2) is 10 m/s upward.

### Step 3: Calculate the change in velocity
The change in velocity (\( \Delta v \)) during the contact with the floor can be calculated as:

\[ \Delta v = v2 - (-v1) \]

Since v1 is downward, we consider it as negative:

\[ \Delta v = 10 - (-20) = 10 + 20 = 30 \, \text{m/s} \]

### Step 4: Calculate the average acceleration
The average acceleration (\( a \)) can be calculated using the formula:

\[ a = \frac{\Delta v}{\Delta t} \]

Where \( \Delta t \) is the time of contact (0.01 s):

\[ a = \frac{30}{0.01} = 3000 \, \text{m/s}^2 \]

### Final Answer
The average acceleration of the tennis ball during contact with the floor is **3000 m/s²**.

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