Two balls A and B are placed at the top of 180 m tall tower. Ball A is released from the top at $t = 0 s$ . Ball B is thrown vertically down with an initial velocity 'u' at $t = 2 s$ . After a certain time, both balls meet 100 m above the ground. Find the value of 'u' in ${ms}^{- 1}$ . (use $g = 10 {ms}^{- 2})$

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answer is D.

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Detailed Solution

Let us assume both ball meet at t=t Sec at height

For 'A'

$h = \frac{1}{2} {gt}^{2} - 80 = \frac{- 1}{2} \times 10 \times t^{2}; t^{2} = \frac{160}{10} \Rightarrow t^{2} = 16, t^{2} = 4 Sec; For ball' B' h = u (t - 2) - \frac{1}{2} g {(t - 2)}^{2}; - 80 = - 2 u - 20; \Rightarrow 2 u = 60 u = 30 m / s$

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