A particle is projected  with a velocity u at an angle thyta with the horizontal . find the radius  of the  parabola traced out by the particle  at the point  where velocity makes an angle thyta/2 with the horizontal.

Dear Student,

Please find below the solution to the asked query:

Given that the initial velocity of the particle is U. The object is projected at an angle θ. Then the horizontal and vertical components of the initial velocity are,

Ux=U cos θ&Uy=U sin θ

Let after t time of projection, the particle is making an angle θ2. Therefore,

V cos θ2=U cos θV=U cos θcos θ2V sin θ2=U cos θcos θ2×sin θ2V=U cos θcos θ22+U cos θcos θ2×sin θ22V=U cos θcos θ21+sin2 θ2

Therefore, the radius of curvature is,

g cos θ=V2RR=V2g cos θ2 R=U cos θcos θ21+sin2θ22g cos θ2R=U2 cos2 θ1+sin2θ2g cos3 θ2

Hope this information will clear your doubts about the topic.

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