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

• A

  $\frac{u^2 \cos ^2 \theta \sec ^3\left(\frac{\theta}{2}\right)}{\sqrt{3} g}$
• B

  $\frac{u^2 \cos ^2 \theta \sec ^3\left(\frac{\theta}{2}\right)}{2 g}$
• C

  $\frac{2 u^2 \cos ^3 \theta \sec ^2\left(\frac{\theta}{2}\right)}{g}$
• Correct answer
  D

  $\frac{u^2 \cos ^2 \theta \sec ^3\left(\frac{\theta}{2}\right)}{g}$

Solution

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, $$ \begin{aligned} \Rightarrow \quad r & =\frac{v^2}{g \cos \frac{\theta}{2}} \\ & =\frac{\left(\cos \frac{\theta}{2}\right)^2}{g \cos \frac{\theta}{2}} \Rightarrow r=\frac{u^2 \cos ^2 \theta \cdot \sec ^3\left(\frac{\theta}{2}\right)}{g} \end{aligned} $$

Hence, the correct answer is option: 'D'.