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what will be the ratio of the distance moved by a freely falling body from rest in 4th and 5th seconds of journey...
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what will be the ratio of the distance moved by a freely falling body ...
Using the equation of motion: s = ut + 1/2gt^2... since its from rest, u=0, thus s = 1/2gt^2 Now the distance moved during the fourth second will be from t = 3 sec to t = 4 sec = 1/2g(4)^2 - 1/2g(3)^2 = 1/2g(7) =7/2 g Similarly, distance moved during fifth second will be from t = 4 sec to t = 5 sec = 1/2g(5)^2 - 1/2g(4)^2 = 1/2g(9) = 9/2 g Thus, the ratio would be 7 : 9
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Most Upvoted Answer
what will be the ratio of the distance moved by a freely falling body ...
7:9 by using this equation,
Hn= vit +1/2 g(2n-1)
as body is freely falling so, initial velocity (vi) is zero, Concluded,
Hn= 1/2 g(2n-1)
where "n" is number of second, and "H" is hight/distance from the ground.
Community Answer
what will be the ratio of the distance moved by a freely falling body ...
7:9 according to galilio's law of odd numbers
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