Two bodies which all are different heights from the ground, start falling under gravity.the Times of fall of the two bodies are 1 sec and 2sec respectively. what is the initial separation between the two bodies?
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Answered by
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H = ut + 1/2gt^2
u = 0, g = 9.8m/s^2, t = 1 or 2s
Height of one body from the surface of the earth
H = 0*1 + 1/2*9.8*1^2
= 0 + 4.9 = 4.9m
Height of the other body from the surface of the earth
H = 0*2 + 1/2*9.8*2^2
0 + 19.6 = 19.6m
Therefore initial separation between the two bodies 19.6-4.9 = 14.7m
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u = 0, g = 9.8m/s^2, t = 1 or 2s
Height of one body from the surface of the earth
H = 0*1 + 1/2*9.8*1^2
= 0 + 4.9 = 4.9m
Height of the other body from the surface of the earth
H = 0*2 + 1/2*9.8*2^2
0 + 19.6 = 19.6m
Therefore initial separation between the two bodies 19.6-4.9 = 14.7m
Now plzz mark me as brainliest
Answered by
6
Two bodies which all are different heights from the ground, start falling under gravity. the Times of fall of the two bodies are 1 sec and 2 sec respectively.
We have to find the initial separation between the two bodies.
- initial velocity of each body is 0 m/s [ because both are falling from certain height. ]
using formula, s = ut + 1/2 at²
here . u = 0 . a = g = 9.8 m/s²
height of the first body = distance covered by the body in 1s , s₁ = 1/2 × 9.8 × 1²
= 4.9 m
height of the second body = distance covered by the body in 2 sec , s₂ = 1/2 × 9.8 × 2²
= 9.8 × 2 = 19.6 m
∴ initial separation between the two bodies = s₂ - s₁ = 19.6 - 4.9 = 14.7 m
Therefore the initial separation between the two bodies is 14.7 m.
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