A body is projected with a speed u at an angl | vedclass

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(D) Let the velocity of the projectile be $v$ at an angle $\frac{	heta}{2}$ with the horizontal.Since the horizontal component of velocity remains co

Vedclass · Accepted Answer

$\frac{u^2 \cos ^2 	heta \sec ^3\left(\frac{	heta}{2}ight)}{g}$

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(D) Let the velocity of the projectile be $v$ at an angle $\frac{	heta}{2}$ with the horizontal.Since the horizontal component of velocity remains constant throughout the motion:$v \cos \left(\frac{	heta}{2}ight) = u \cos 	heta$$v = \frac{u \cos 	heta}{\cos \left(\frac{	heta}{2}ight)}$The centripetal force required for the curved path is provided by the component of gravity perpendicular to the velocity vector,which is $mg \cos \left(\frac{	heta}{2}ight)$.Using the formula for centripetal force $\frac{mv^2}{r} = F_c$:$\frac{mv^2}{r} = mg \cos \left(\frac{	heta}{2}ight)$$r = \frac{v^2}{g \cos \left(\frac{	heta}{2}ight)}$Substituting the value of $v$:$r = \frac{\left(\frac{u \cos 	heta}{\cos \left(\frac{	heta}{2}ight)}ight)^2}{g \cos \left(\frac{	heta}{2}ight)}$$r = \frac{u^2 \cos^2 	heta}{g \cos^2 \left(\frac{	heta}{2}ight) \cdot \cos \left(\frac{	heta}{2}ight)}$$r = \frac{u^2 \cos^2 	heta \sec^3 \left(\frac{	heta}{2}ight)}{g}$

$A$ 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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