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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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Solution 1

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.

∵ H2>H1

⟹ v2>v1

∴ t2<t1

Hence Option 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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