A body is thrown up with a velocity `40 ms ^(-1).` At same time another body is dropped from a height 40 m. Their relative acceleration after `1.3` seconds is

Options

A. 4g
B. g/2
C. 2g
D. zero

Answer

To find the relative acceleration of the two bodies after 1.3 seconds, we can follow these steps:

### Step 1: Identify the accelerations of both bodies
- The body thrown upwards (let's call it Body B) has an initial velocity of \( 40 \, \text{m/s} \) and is subject to gravitational acceleration \( g \) acting downwards.
- The body dropped from a height (let's call it Body A) is also subject to gravitational acceleration \( g \) acting downwards.

### Step 2: Write down the acceleration of each body
- The acceleration of Body A (dropped) is \( g \) downwards.
- The acceleration of Body B (thrown upwards) is also \( g \) downwards.

### Step 3: Calculate the relative acceleration
- The relative acceleration of Body A with respect to Body B can be calculated using the formula:
\[ a_{AB} = a_A - a_B \]
- Since both bodies have the same acceleration \( g \) downwards, we have:
\[ a_{AB} = g - g = 0 \]

### Step 4: Conclusion
- The relative acceleration of the two bodies after \( 1.3 \) seconds is \( 0 \, \text{m/s}^2 \).

### Final Answer
The relative acceleration after \( 1.3 \) seconds is \( 0 \, \text{m/s}^2 \).

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