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A ball is dropped from a high rise platform at t = 0 starting from rest. After 6 s another ball is thrown downwards from the same platform with a speed v. The two balls meet at t = 18 s. What is the value of v?(take g = 10 ${ms}^{- 2}$ )

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75 ${ms}^{- 1}$

55 ${ms}^{- 1}$

40 ${ms}^{- 1}$

60 ${ms}^{- 1}$

answer is A.

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Detailed Solution

For first ball, u = 0

$∴$ $s_{1} = \frac{1}{2} {gt}_{1}^{2} = \frac{1}{2} \times g \times 18^{2}$

For second ball, initial velocity = v

$∴$ $s_{2} = {vt}_{2} + \frac{1}{2} {gt}_{2}^{2}$

$t_{2} = 18 - 6 = 12 s$

$\Rightarrow s_{2} = s_{2} = v \times 12 + \frac{1}{2} g {(12)}^{2}$

Here, $s_{1} = s_{2}$

$\frac{1}{2} g {(18)}^{2} = 12 v + \frac{1}{2} g {(12)}^{2} \Rightarrow v = 75 {ms}^{- 1}$

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