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The ratio of the distances travelled by a freely falling body in the 1st, 2nd, 3rd and 4th second : Acceleration due to gravity=g;time elapsed=t;displacement=s. To find the ratio of distances travelled by a freely falling body in the 2nd and 3rd seconds, we can use the formula for distance travelled in free fall.
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S = u + 1 2 a (2 t 1) s is the distance travelled in t th second. If an object is moving with constant acceleration 'a', initial velocity 'u', the distance travelled in tth second is given by: Now to calculate the ratio of the distances traveled by a freely falling body in the 1st, 2nd, 3rd, and 4th second we divide the equations (3), (4), (5), and (6) we get;
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S = 1 2 g t 2 ; The distance travelled by freely falling body, initial velocity=u=0; The ratio of the distances travelled by a freely falling body in the 1 text st , 2 text nd , 3 text rd and 4 text th second:
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The ratio of distance is g 2: Distance travelled in third second i.e. 5 g 2 = 1:
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Distance travelled in 2nd second i.e. Calculate the ratio of distances travelled by a body falling freely from rest in the first, second, and third second using kinematic equations.
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