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Q.

A ball is dropped from a high rise platform at t = 0 starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed u. The two balls meet at t =18s. What is the value of v ? (take g = 10m/s2)

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a

60 m/s

b

7 5 m/s

c

55 m/s

d

40 m/s

answer is B.

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Detailed Solution

Let two balls meet at depth h from platform, so 
h=12g182=v12+12g122⇒v=75m/s

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