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Answer
As per the per the law of motion v^2 = u^2 - 2gs S= ut +1/2 at^2 S= 9 G= 10 ms^-2 S T =✓2s /g ✓2*9/10 1.34 s Horizontal distance covered = it = 9*1.34.>10 m So the person would land safely
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Yes, the man will be able to land on the next building...
Answer

H = u x t + 0.5 x g x t x t

Here u=0 (velocity downward) and Heiught Diff is 9 , 9 = 0.5 x 10 x t ^ 2  So, t = Sqrt(9/5) --> This is the maximum time by which he should land on the other roof Now v = 9m/s and t= 3 /Sqrt(5) = 3/2.23 = 1.34s Distance Travelled = 9 x 1.34 = 12m > 10m (the distance between buildings) So, the answer is YES!

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