A particle projected from ground withspeed u atan angle q with the horizontal. Radius of curvature of trajectory ofthe particle

Dear student,

Equation of Trajectory (Path of projectile)

At any instant t
x= ucosØ.t
» t= x/(ucosØ)

Also , y= usinØ.t - (1/2)gt²

Substituting for t
y= usinØ.x/(ucosØ) - (1/2)g[x/(ucosØ)]²

» y= x.tanØ - [(1/2)g.sec²Ø.x² ]/u²

This equation is of the form y= ax + bx² where 'a' and 'b are constants.This is the equation of a parabola.Thus,the path of a projectile is a parabola .

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Dear student,

Please do not post the same doubt again n again...

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