A ball dropped on to the floor from a height of 40 m and rebounds to a height of 10 m. If the ball is in contact with the floor for 0.02 s, its average acceleration during contact is

Options

A. `1050 "m/s"^2`
B. `2100 "m/s"^2`
C. `9.8 "m/s"^2`
D. `210 "m/s"^2`

Answer

To solve the problem of finding the average acceleration of the ball during its contact with the floor, we can follow these steps:

### Step 1: Identify the initial and final velocities
- The ball is dropped from a height of 40 m. When it reaches the floor, its initial velocity (u) just before impact can be calculated using the formula for free fall:
\[ v^2 = u^2 + 2gh \]
Here, \(u = 0\) (the ball is dropped), \(g = 9.8 \, \text{m/s}^2\), and \(h = 40 \, \text{m}\).
Thus, we can calculate:
\[ v^2 = 0 + 2 \times 9.8 \times 40 \]
\[ v^2 = 784 \implies v = \sqrt{784} = 28 \, \text{m/s} \]
Therefore, just before hitting the ground, the velocity \(u = 28 \, \text{m/s}\) (downward).

### Step 2: Calculate the final velocity after rebound
- After rebounding to a height of 10 m, we can find the final velocity (v) just after the rebound using the same formula:
\[ v^2 = u^2 + 2gh \]
Here, the initial velocity just after the rebound is \(u = 0\) (as it momentarily stops at the peak), and \(h = 10 \, \text{m}\):
\[ v^2 = 0 + 2 \times 9.8 \times 10 \]
\[ v^2 = 196 \implies v = \sqrt{196} = 14 \, \text{m/s} \]
Thus, the final velocity after the rebound is \(v = 14 \, \text{m/s}\) (upward).

### Step 3: Calculate the average acceleration
- The average acceleration \(a\) during the contact time with the floor can be calculated using the formula:
\[ a = \frac{v - u}{\Delta t} \]
Here, \(v = 14 \, \text{m/s}\) (upward) and \(u = -28 \, \text{m/s}\) (downward, hence negative):
\[ a = \frac{14 - (-28)}{0.02} \]
\[ a = \frac{14 + 28}{0.02} = \frac{42}{0.02} = 2100 \, \text{m/s}^2 \]

### Final Answer:
The average acceleration of the ball during contact with the floor is \(2100 \, \text{m/s}^2\).

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