A ball dropped from one metre above the top of a window, crosses the window in t1s . If the same ball is dropped from 2m above the top of the same window, time taken by it to cross the window is t2s . Then
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The ball which is dropped from greater height above the top of the window will have higher speed while crossing the window, by 3rd equation of motion:
v2=u2+2aS, where u=0
The ball with higher speed will take less time to cross the window.
Hence Option D is correct
Statement D is correct Situation 1: We want time to cross x. v2=u2+2as u=0,a=−g,s=−1 v2=2g⇒v1=√2g (1+x)=12gt2⇒T2=√2(1+x)g 1=12gt2=T1=√2g So time taken to cross x=t1=T2−T1=√2g(√1+x−1) Situation 2: (2+x)=12gt2⇒T2=√2(2+x)g 2=12gt2⇒T1=√2×2g So time taken to cross x=t2=T2−T1=√2g(√2+x−√2) Clearly t2<t1
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