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A stone is thrown in such a manner that it would just hit a bird at the top of a tree and afterwards reach a maximum height double that of the tree. If at the moment of throwing the stone the bird flies away horizontally with constant velocity ant the stone hits the bird after some time. The ratio of horizontal velocity of stone to that of the bird is 1/n +1/√n. Find 2n.
Dear Mithun For the stone: uy = vertical speed ux = horizontal speedLet the height of the tree = h max height reached = 2h=> uy^2 = 2 (2h) g = 4 h g --- (1)Let the speed of the bird be v.The stone and bird meet at height y = h, two times t1 and t2.y =uy t - 1/2 g t² => h = 2√(hg) t - 1/2 g t²=> t1 = √(2h/g) * [√2 - 1] t2 = √(2 h/g) * [√2 + 1 ]at t = t1, the stone travels a distance = ux * t1at t = t2, the stone travels a distance = ux * t2 = ux * t1 + v * t2so ux / v = t2/(t2 - t1) = (√2 + 1)/2 = 1/√2 + 1/2So n = 2 and 2n = 4 RegardsArun
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