A particle is projected from a point A with a velocity at an angle wand) with the horizontal At a
certain point moves a night angle to its initialecto follows that
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Answer:
Horizontal component of velocity remains
unchanged
∴vcosθ=v'cos(90−θ)
or v'=vcotθ
In vertical (y) direction,
vy=uy+ayt
∴t=vy−uyay
=−v1sin(90−θ)−vsinθ−g
=(vcotθ)⋅cosθ+vsinθg
=vcosecθg .
Explanation:
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