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A body released from a great height falls freely towards earth. Another body is released from the same height exactly one second later. The separation between the two bodies two seconds after the release of the second body is
- a)19.6 m
- b)24.5 m
- c)4.9 m
- d)9.8 m
Correct answer is option 'B'. Can you explain this answer?
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A body released from a great height falls freely towards earth. Anothe...
Solution:
Given: The two bodies are released from the same height, and the second body is released one second later.
Let us consider the motion of the two bodies separately.
For the first body:
Initial velocity, u = 0 (since it is released from rest)
Acceleration due to gravity, a = 9.8 m/s² (since it is falling freely towards earth)
Time, t = 2 seconds (since we need to find the separation after two seconds)
Using the formula for distance covered by a body under constant acceleration:
S = ut + ½at²
S = 0 + ½(9.8)(2) = 9.8 m
Therefore, the first body falls a distance of 9.8 meters in two seconds.
For the second body:
Initial velocity, u = 0 (since it is released from rest)
Acceleration due to gravity, a = 9.8 m/s² (since it is falling freely towards earth)
Time, t = 1 second (since it is released one second after the first body)
Using the same formula as before:
S = ut + ½at²
S = 0 + ½(9.8)(1) = 4.9 m
Therefore, the second body falls a distance of 4.9 meters in one second.
Now, to find the separation between the two bodies after the release of the second body, we need to subtract the distance covered by the second body from the distance covered by the first body.
Separation = Distance covered by first body - Distance covered by second body
Separation = 9.8 - 4.9 = 4.9 m
Therefore, the separation between the two bodies two seconds after the release of the second body is 4.9 meters.
Hence, the correct answer is option (B) 4.9 m.
Given: The two bodies are released from the same height, and the second body is released one second later.
Let us consider the motion of the two bodies separately.
For the first body:
Initial velocity, u = 0 (since it is released from rest)
Acceleration due to gravity, a = 9.8 m/s² (since it is falling freely towards earth)
Time, t = 2 seconds (since we need to find the separation after two seconds)
Using the formula for distance covered by a body under constant acceleration:
S = ut + ½at²
S = 0 + ½(9.8)(2) = 9.8 m
Therefore, the first body falls a distance of 9.8 meters in two seconds.
For the second body:
Initial velocity, u = 0 (since it is released from rest)
Acceleration due to gravity, a = 9.8 m/s² (since it is falling freely towards earth)
Time, t = 1 second (since it is released one second after the first body)
Using the same formula as before:
S = ut + ½at²
S = 0 + ½(9.8)(1) = 4.9 m
Therefore, the second body falls a distance of 4.9 meters in one second.
Now, to find the separation between the two bodies after the release of the second body, we need to subtract the distance covered by the second body from the distance covered by the first body.
Separation = Distance covered by first body - Distance covered by second body
Separation = 9.8 - 4.9 = 4.9 m
Therefore, the separation between the two bodies two seconds after the release of the second body is 4.9 meters.
Hence, the correct answer is option (B) 4.9 m.
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Community Answer
A body released from a great height falls freely towards earth. Anothe...
Calculate the distance covered by the first body in 3 seconds and second body in 2 seconds . difference comes out to be 24.5m
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A body released from a great height falls freely towards earth. Another body is released from the same height exactly one second later. The separation between the two bodies two seconds after the release of the second body isa)19.6 mb)24.5 mc)4.9 md)9.8 mCorrect answer is option 'B'. Can you explain this answer?
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