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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 v. 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

40 m/s

d

55 m/s

answer is B.

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Detailed Solution

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Let two balls meet at depth h from platform, so
$h = \frac{1}{2} g {(18)}^{2} = v (12) + \frac{1}{2} g {(12)}^{2} \Rightarrow v = 75 m / s$

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