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. `2100 m// sec^(2) ` downwards
B. ` 2100m//sec^(2) ` upwards
C. ` 1400m //sec ^(2) ` upwards
D. ` 700 m// sec ^(2) ` upwards

Explanation

To solve the problem step by step, we need to find the average acceleration of the ball while it is in contact with the floor. Here's how we can do that:

### Step 1: Determine the velocity just before impact
The ball is dropped from a height of 10 m. We can use the equation of motion to find the velocity just before it hits the ground:

\[ V^2 = U^2 + 2gH \]

Where:
- \( V \) = final velocity just before impact
- \( U \) = initial velocity (0 m/s, since it is dropped)
- \( g \) = acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \))
- \( H \) = height (10 m)

Substituting the values:

\[ V^2 = 0 + 2 \times 9.8 \times 10 \]
\[ V^2 = 196 \]
\[ V = \sqrt{196} = 14 \, \text{m/s} \]

### Step 2: Determine the velocity just after rebound
The ball rebounds to a height of 2.5 m. We will again use the equation of motion to find the velocity just after it leaves the floor:

\[ U^2 = V^2 + 2gH \]

Where:
- \( U \) = initial velocity just after the rebound (what we want to find)
- \( V \) = final velocity at the peak of the rebound (0 m/s, since it momentarily stops at the peak)
- \( g \) = \( 9.8 \, \text{m/s}^2 \)
- \( H \) = height (2.5 m)

Substituting the values:

\[ 0 = U^2 - 2 \times 9.8 \times 2.5 \]
\[ U^2 = 2 \times 9.8 \times 2.5 \]
\[ U^2 = 49 \]
\[ U = \sqrt{49} = 7 \, \text{m/s} \]

### Step 3: Calculate the average acceleration during contact
Now we can calculate the average acceleration while the ball is in contact with the floor. The average acceleration \( a \) is given by:

\[ a = \frac{V_f - V_i}{t} \]

Where:
- \( V_f \) = final velocity after rebound (7 m/s, upward)
- \( V_i \) = initial velocity just before impact (14 m/s, downward, so we take it as -14 m/s)
- \( t \) = time of contact (0.01 s)

Substituting the values:

\[ a = \frac{7 - (-14)}{0.01} \]
\[ a = \frac{7 + 14}{0.01} \]
\[ a = \frac{21}{0.01} \]
\[ a = 2100 \, \text{m/s}^2 \]

### Conclusion
The average acceleration during contact with the floor is approximately \( 2100 \, \text{m/s}^2 \) in the upward direction.

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