Arun
Last Activity: 6 Years ago
Dear Mithun
For the stone: uy = vertical speed ux = horizontal speed
Let 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 * t1
at t = t2, the stone travels a distance = ux * t2 = ux * t1 + v * t2
so ux / v = t2/(t2 - t1) = (√2 + 1)/2 = 1/√2 + 1/2
So n = 2 and 2n = 4
Regards
Arun