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Question

A ball is dropped from the top of a building 100m high. At the same instant another ball is thrown upwards with a velocity of 40ms1 from the building. The two balls will meet after.

A
5s
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B
2.5s
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C
2s
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D
3s
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Solution

The correct option is A 2.5s
Given,

H=100m

For ball 1: u=0;g=10m/s

So, h1=u1+12gt2=0×t+12(10)2t2=5t2 ----------(1)

For ball2: u=40m/s;a=10m/s

So, h2=u2+12at2=40×t+12(10)2t2=40t5t2 ---------(2)

h=h1+h2

h=5t2+40t5t2h=100m

100=40t

t=52=2.5s

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