Arun
Last Activity: 5 Years ago
Let ‘2m’ be the mass of the cannon which is projected with velocity ‘v’ at an angle ‘θ’ with the horizontal. The horizontal component of the velocity of the cannon is, vx = v cosθ. This component remains constant.
At the maximum height, the cannon has only this component as the vertical component is momentarily zero.
So, momentum before exploding = 2mv cosθ
After exploding, one of the masses retraces the path of the cannon backwards. This is possible on if the part has velocity ‘-v cosθ’.
Therefore, momentum after explosion = -mv cosθ + mu
So, by conservation of momentum,
2mv cosθ = -mv cosθ + mu
= > u = (3v cosθ) m/s
Hope it helps
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