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Question

From the top of a tower a particle is thrown down with velocity 10m/s. Find ratio of distances travelled by it in third and second second 


Solution

ANSWER :-
the formula for distance covered in nth second
Sn = u + a/2 * ( 2n - 1)
S3 = 10 + 5 * 5 = 35 m Here a = g = 10m s^-2 (for covenience)
S2 = 10 + 5 * 3 = 25 m
So ratio required = 35/25 = 7/5
7:5 is the required ratio

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