A ball is dropped on the floor from a height of 10m. It rebounds to a height of 2.5 m if the ball is in contact with floor for 0.01 s then the average acceleration during contact is nearly

Options

A. `700 ms^(-2)`
B. `1400ms^(-2)`
C. `2100 ms^(-2)`
D. `2800 ms^(-2)`

Explanation

To solve the problem step by step, we will calculate the average acceleration of the ball during its contact with the floor.

### Step 1: Calculate the velocity just before the ball hits the ground (v1)
The ball is dropped from a height of 10 m. We can use the following kinematic equation to find the velocity just before it hits the ground:

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

Where:
- \( v \) = final velocity (just before hitting the ground)
- \( u \) = initial velocity (0 m/s, since it is dropped)
- \( a \) = acceleration due to gravity (\( -g = -10 \, \text{m/s}^2 \))
- \( s \) = displacement (height = -10 m)

Substituting the values:

\[ v_1^2 = 0 + 2 \times (-10) \times (-10) \]
\[ v_1^2 = 200 \]
\[ v_1 = \sqrt{200} = 10\sqrt{2} \, \text{m/s} \]

### Step 2: Calculate the velocity just after the ball rebounds (v2)
The ball rebounds to a height of 2.5 m. We can use the same kinematic equation to find the velocity just after it leaves the ground:

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

Where:
- \( v \) = final velocity (0 m/s at the maximum height)
- \( u \) = initial velocity (just after rebounding)
- \( a \) = acceleration due to gravity (\( -g = -10 \, \text{m/s}^2 \))
- \( s \) = displacement (height = +2.5 m)

Substituting the values:

\[ 0 = u_2^2 + 2 \times (-10) \times 2.5 \]
\[ u_2^2 = 50 \]
\[ u_2 = \sqrt{50} = 5\sqrt{2} \, \text{m/s} \]

### Step 3: Calculate the average acceleration during contact
The average acceleration (\( a_{avg} \)) can be calculated using the formula:

\[ a_{avg} = \frac{\Delta v}{\Delta t} \]

Where:
- \( \Delta v = v_2 - v_1 \)
- \( \Delta t = 0.01 \, \text{s} \)

Substituting the values:

\[ \Delta v = 5\sqrt{2} - (-10\sqrt{2}) = 5\sqrt{2} + 10\sqrt{2} = 15\sqrt{2} \, \text{m/s} \]

Now substituting into the average acceleration formula:

\[ a_{avg} = \frac{15\sqrt{2}}{0.01} \]
\[ a_{avg} = 1500\sqrt{2} \, \text{m/s}^2 \]

Calculating \( 1500\sqrt{2} \):

\[ \sqrt{2} \approx 1.414 \]
\[ a_{avg} \approx 1500 \times 1.414 \approx 2121 \, \text{m/s}^2 \]

### Final Answer
The average acceleration during contact is approximately \( 2100 \, \text{m/s}^2 \).

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