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Three stones A, B and C are simultaneously projected from same point with same speed. A is thrown upwards, B is thrown horizontally and C is thrown downwards from a building. When the distance between stone A and C becomes 10 m, then distance between A and B will be :

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$10 \sqrt{2} m$

5 m

10 m

$5 \sqrt{2} m$

answer is C.

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Detailed Solution

$S_{1} = - U t + \frac{1}{2} g t^{2} S_{2} = U t + \frac{1}{2} g t^{2} \begin{array}{l} \Rightarrow 10 = 2 U t \\ U t = 5 \end{array}$

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