A ball dropped on to the floor from a height of 10 m rebounds to a height of 2.5 m. If the ball is in contact with the floor for 0.02s, its average acceleration during contact is

Options

A. 2100 `"ms"^(-2)`
B. 1050 `"ms"^(-2)`
C. 4200 `"ms"^(-2)`
D. 9.8 `"ms"^(-2)`

Answer

To solve the problem, we need to find the average acceleration of the ball during the time it is in contact with the floor. We will follow these steps:

### Step 1: Calculate the velocity just before the ball hits the ground (u1)
We can use the third equation of motion:
\[ v^2 = u^2 + 2as \]
Where:
- \( v \) = final velocity (just before hitting the ground)
- \( u \) = initial velocity (0 m/s, since the ball is dropped)
- \( a \) = acceleration (which is \( g \), approximately \( 9.8 \, \text{m/s}^2 \))
- \( s \) = distance fallen (10 m)

Substituting the values:
\[ v^2 = 0 + 2 \cdot 9.8 \cdot 10 \]
\[ v^2 = 196 \]
\[ v = \sqrt{196} = 14 \, \text{m/s} \]
So, the velocity just before the ball hits the ground (u1) is \( 14 \, \text{m/s} \).

### Step 2: Calculate the velocity just after the ball rebounds (u2)
We can use the same equation of motion for the rebound height of 2.5 m:
\[ v^2 = u^2 + 2as \]
Where:
- \( v \) = final velocity (0 m/s, at the peak of the rebound)
- \( u \) = initial velocity (which we need to find)
- \( a \) = acceleration (which is \( -g \), since it is moving upwards)
- \( s \) = distance risen (2.5 m)

Substituting the values:
\[ 0 = u^2 - 2 \cdot 9.8 \cdot 2.5 \]
\[ u^2 = 2 \cdot 9.8 \cdot 2.5 \]
\[ u^2 = 49 \]
\[ u = \sqrt{49} = 7 \, \text{m/s} \]
So, the velocity just after the ball rebounds (u2) is \( 7 \, \text{m/s} \).

### Step 3: Calculate the average acceleration during contact with the floor
The average acceleration \( a \) can be calculated using the formula:
\[ a = \frac{v_f - v_i}{\Delta t} \]
Where:
- \( v_f \) = final velocity (upward, \( 7 \, \text{m/s} \))
- \( v_i \) = initial velocity (downward, \( -14 \, \text{m/s} \))
- \( \Delta t \) = time of contact (0.02 s)

Substituting the values:
\[ a = \frac{7 - (-14)}{0.02} \]
\[ a = \frac{7 + 14}{0.02} \]
\[ a = \frac{21}{0.02} \]
\[ a = 1050 \, \text{m/s}^2 \]

### Final Answer
The average acceleration of the ball during contact with the floor is \( 1050 \, \text{m/s}^2 \).

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