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A body is thrown up with a velocity 40 ...

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

A

4g

B

g/2

C

2g

D

zero

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the motion of both bodies and determine their relative acceleration after 1.3 seconds. ### Step-by-Step Solution: 1. **Identify the motions of the two bodies**: - **Body A**: Thrown upwards with an initial velocity of \( u_A = 40 \, \text{m/s} \). - **Body B**: Dropped from a height of \( h = 40 \, \text{m} \) with an initial velocity of \( u_B = 0 \, \text{m/s} \). 2. **Determine the acceleration of both bodies**: - Both bodies experience the same acceleration due to gravity, \( g \approx 9.81 \, \text{m/s}^2 \), acting downwards. - Therefore, the acceleration of Body A, \( a_A = -g \) (upward motion is considered negative) and for Body B, \( a_B = -g \). 3. **Calculate the relative acceleration**: - The relative acceleration \( a_{rel} \) between two bodies is given by: \[ a_{rel} = a_A - a_B \] - Substituting the values: \[ a_{rel} = (-g) - (-g) = -g + g = 0 \] 4. **Conclusion**: - After 1.3 seconds, the relative acceleration of the two bodies 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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