A body is projected with a speed $u$ at an angle $ heta$ with the horizontal. The radius of curvature of the trajectory, when it makes an angle $\left(\frac{ heta}{2} ight)$ with the horizontal is ( $g$-acceleration due to gravity)

Question

A body is projected with a speed $u$ at an angle $	heta$ with the horizontal. The radius of curvature of the trajectory, when it makes an angle $\left(\frac{	heta}{2}ight)$ with the horizontal is ( $g$-acceleration due to gravity)

Aicharya · Accepted Answer

Let velocity of projectile is $v$ at an angle $\frac{\theta}{2}$ with horizontal $ \begin{aligned} \therefore & v \cos \frac{\theta}{2}=u \cos \theta \\ \text { or } \quad v & =\frac{u \cos \theta}{\cos \frac{\theta}{2}} \end{aligned} $

As horizontal component remains same. Also, centripetal force is provided by the component of weight. So, $ \frac{m v^2}{r}=m g \cos \frac{\theta}{2} $ Hence, radius of curvature of path, $ \begin{aligned} \Rightarrow \quad r & =\frac{v^2}{g \cos \frac{\theta}{2}} \\ & =\frac{\left(\cos \frac{\theta}{2}\right)^2}{g \cos \frac{\theta}{2}} \Rightarrow r=\frac{u^2 \cos ^2 \theta \cdot \sec ^3\left(\frac{\theta}{2}\right)}{g} \end{aligned} $

A body is projected with a speed $u$ at an angle $\theta$ with the horizontal. The radius of curvature of…

Solution