Given, Let height of ball from ground be H and time taken by ball l to reach on ground be t. ∴H=500m ∴ Time taken by ball 2 to reach on ground t2=(t−1) As we know that H=ut+
1
2
gt2 where, u is initial speed i.e. 0ms−1 for free fall and g is acceleration due to gravity i.e. 10ms−2. ∴500=0t+
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gt2 [For the first ball] ⇒500=
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×10t2 ⇒100=t2 ⇒t=10s ∴H2 (distance covered by ball 2) H2=u(t−1)+
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g(t−1)2 ⇒H2=
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g(10−1)2 [∵u=0] =
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2
×10×92 =5×81=405m Hence, distance between the two balls, when first ball is in contact with ground is equal to ∆H, then ∆H=H1−H2 =500−405=95m
