CBSE Class 11-science Answered
show that for a freely falling body
the ratio of distances traversed after 1second 2 second and 3 second are in the ratio 1square:2square:3square
the velocities are in 1:2:3
Asked by sonu300502 | 29 Jun, 2019, 05:22: PM
Expert Answer
For a freely falling body, vertical distance S travelled in a time t seconds, S = (1/2)g×t2 ................(1)
If S1 , S2 and S3 are the respective distances travelled after 1 sec, 2 sec and 3 sec,
then their ratio = (1/2)g×12 : (1/2)g×22 : (1/2)g×32 = 12 : 22 : 32
For a freely falling body, velocity after t seconds, v = g×t ................(2)
If v1 , v2 and v3 are the respective velocities after 1 sec, 2 sec and 3 sec,
then their ratio = g×1 : g×2 : g×3 = 1 : 2 : 3
Answered by Thiyagarajan K | 30 Jun, 2019, 10:29: AM
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