Question

What is the radius of curvature of the parabola traced by a projectile with initial speed $u$ at an angle of $2\theta$ with the horizontal, at a point where the velocity of the particle makes an angle $\theta$ with the horizontal?

• A. $\frac{u^2{\cos }^{2}2\theta }{g{\cos }^3\theta }$
  
• B. $\frac{u^2{\cos }^{2}2\theta }{g{\cos }^2\theta }$
  
• C. $\frac{u^2{\cos }^{3}2\theta }{g{\cos }^2\theta }$
  
• D. $\frac{u^2\cos 2\theta }{g{\cos }^3\theta }$
  

Answer 1

The correct option is A $\frac{u^2{\cos }^{2}2\theta }{g{\cos }^3\theta }$


From the figure shown below:


$u_x=u\cos 2\theta ,v_x=v_2\cos \theta$
Horizontal component of the velocity do not changes, because $a_x=0$

$v_x=u_x$
$\Rightarrow v_2=\frac{u\cos 2\theta }{\cos \theta }$

Now component of acceleration along radial direction, for circle of radius of curvature $R$ is a part & $v_2$ is the tangential speed.

$a_n=g\cos \theta$.......(i)
But we know $a_n=\frac{v_2^2}{R}$........(ii)

From Eq.(i) and Eq.(ii):
$R=\frac{v_2^2}{a_n}=\frac{\frac{u^2{\cos }^{2}2\theta }{{\cos }^2\theta }}{g\cos \theta }\therefore R=\frac{u^2{\cos }^{2}2\theta }{g{\cos }^3\theta }$