At t = 0 a bird starts flying horizontally from the top of a tower of a height 1.25 m with initial acceleration of 1.5 m/s2. At the same instant a stone is projected from point B on ground as shown. If stone hits the bird during its downward motion at an angle 45° to the horizontal, the time at which stone will hit the bird:
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for bird x = ½ at2 = 0.75 t2
for stone 3 + x = u cosθ t
also in 'y' 1.25 = usinθ × t – ½ gt2
solving
4.25 + 0.75 t2 = (usinθ + ucosθ)t – 5t2 ...(1)
At the point of hitting vx = vy v cosθ
= – (u sinθ –gt) for – ve direction
(u cosθ + usinθ)] = gt
Solving (1)
4.25 + 0.75 t2 = gt2 – 5t2
4.25 + 0.75 t2 = 5t2
t = 1 sec.