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A particle is projected from a pipe at an angle \(\theta = 45^o\) wtih the vertical at a point P with a speed \(u = 10 ms^{-1}\). A strong horizontal wind gives a constant horizontal acceleration equal to \(5 ms^{-2}\) to the particle such that the particle reaches the ground at point Q which is exactly below P. The height of point P (in m) from the ground is \([g = 10 ms^{-2}]\) 

(a) 30

(b) 40

(c) 50

(d) 60


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Asked By Sudhanva gv

Updated Thu, 20 Sep 2018 04:33 pm

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