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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)
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Verified Answer
The correct answer is:
$\frac{u^2 \cos ^2 \theta \sec ^3\left(\frac{\theta}{2}\right)}{g}$
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,
$$
\begin{aligned}
\Rightarrow \quad r & =\frac{v^2}{g \cos \frac{\theta}{2}} \\
& =\frac{\left(\cos \frac{\theta}{2}\right)^2}{g \cos \frac{\theta}{2}} \Rightarrow r=\frac{u^2 \cos ^2 \theta \cdot \sec ^3\left(\frac{\theta}{2}\right)}{g}
\end{aligned}
$$
$$
\begin{aligned}
\therefore & v \cos \frac{\theta}{2}=u \cos \theta \\
\text { or } \quad v & =\frac{u \cos \theta}{\cos \frac{\theta}{2}}
\end{aligned}
$$
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,
$$
\begin{aligned}
\Rightarrow \quad r & =\frac{v^2}{g \cos \frac{\theta}{2}} \\
& =\frac{\left(\cos \frac{\theta}{2}\right)^2}{g \cos \frac{\theta}{2}} \Rightarrow r=\frac{u^2 \cos ^2 \theta \cdot \sec ^3\left(\frac{\theta}{2}\right)}{g}
\end{aligned}
$$
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