Question

# 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

A
t2=t1
B
t2=2t1
C
t2>t1
D
t2<t1
Solution
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#### Solution 1The 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=0The ball with higher speed will take less time to cross the window.∵ H2>H1 ⟹ v2>v1 ∴ t2<t1Hence Option D is correctSolution 2Statement D is correctSituation 1:We want time to cross x.v2=u2+2asu=0,a=−g,s=−1v2=2g⇒v1=√2g(1+x)=12gt2⇒T2=√2(1+x)g1=12gt2=T1=√2gSo time taken to cross x=t1=T2−T1=√2g(√1+x−1)Situation 2:(2+x)=12gt2⇒T2=√2(2+x)g2=12gt2⇒T1=√2×2gSo time taken to cross x=t2=T2−T1=√2g(√2+x−√2)Clearly t2<t1

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