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,

$$\begin{gathered} U_x=U \cos \theta \\ \& \\ {U}_y=U \sin \theta \end{gathered}$$

Let after t time of projection, the particle is making an angle $\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$. Therefore,

$$\begin{gathered} V \cos \left(\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)=U \cos \theta \Rightarrow V=\frac{U cos \theta }{cos \left(\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)} \\ \therefore V \sin \left(\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)=\left(\frac{U cos \theta }{cos \left(\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}\right)\times \sin \left(\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right) \\ \Rightarrow V=\sqrt{{\left(\frac{U cos \theta }{cos \left(\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}\right)}^2+{\left(\left(\frac{U cos \theta }{cos \left(\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}\right)\times sin \left(\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\right)}^2} \\ \Rightarrow V=\frac{U cos \theta }{cos \left(\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}\sqrt{1+si{n}^2 \left(\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)} \end{gathered}$$

Therefore, the radius of curvature is,

$$\begin{gathered} g \cos \theta =\frac{V^2}{R}\Rightarrow R=\frac{V^2}{g \cos \left({\displaystyle \raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)} \Rightarrow R=\frac{{\left[{\displaystyle \frac{U \cos \theta }{\cos \left({\displaystyle \raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}}\sqrt{1+{\sin }^2\left({\displaystyle \raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}\right]}^2}{g cos \left(\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)} \\ \Rightarrow R=\frac{{\displaystyle U^2 co{s}^2 \theta \left(1+si{n}^2\left(\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\right)}}{g co{s}^3 \left(\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)} \end{gathered}$$
 

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