Question

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

Answer 1

Answer: (D)

Let velocity of projectile is $v$ at an angle $\frac{\theta}{2}$ with horizontal
$\therefore v \cos \frac{\theta}{2} =u \cos \theta$
or $v =\frac{u \cos \theta}{\cos \frac{\theta}{2}}$

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