The ratio of time taken to cover $$1st$$ m, $$2nd$$ m, $$3rd$$ m distances in freely falling motion is

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Infinity Learn · Accepted Answer

time taken in free fall $$t=2hgequation$$ of motion, $$S=ut+12gt2$$  initial $$velocity=U$$;acceleration due to $$gravity=g$$;time $$elapsed=t$$;$$displacement=S$$ for freely falling body, $$S=0+12gt2$$  ⇒$$t=2hg$$ time for $$displacement=S=1m$$  is $$ts=1=$$ $$2(1)g=2g$$ time for $$displacement=S=2m$$  is $$ts=2=$$ $$2(2)g=2g2$$ time for $$displacement=S=3m$$  is $$ts=3=$$ $$2(3)g=2g3$$ time for $$displacement=S=4m$$  is $$ts=4=$$ $$2(4)g=2g4$$ time taken to $$1st$$ $$m=T1=$$ $$ts=1-$$ $$ts=0=2g-0=2g$$ time taken to $$cover2nd$$ $$m=T2=$$ $$ts=2-$$ $$ts=1=2g2-2g=2g(2-1)$$ time taken to $$cover3rd$$ $$m=T3=$$ $$ts=3-$$ $$ts=2=2g3-2g2=2g(3-2)$$ time taken to cover  $$4th$$ $$m=T4=$$ $$ts=4-$$ $$ts=3=2g4-2g3=2g(4-3)$$ $$T1$$:$$T2$$:$$T3=2g$$:$$2g(2-1)$$:$$2g(3-2)$$:$$2g(4-3)$$ $$T1$$:$$T2$$:$$T3=1$$:(2-1):(3-2):$$(4-3)=ratio$$ of time taken for successive meter $$displacementt1$$:$$t2$$:$$t3=1$$:$$2$$−$$1$$:$$3$$−$$2$$

The ratio of time taken to cover 1st m, 2nd m, 3rd m distances in freely falling motion is

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