A projectile is thrown with a speed $u$ at an angle $ heta$ to the horizontal. The radius of curvature of its trajectory when the velocity vector of the projectile makes an angle $\alpha$ with the horizontal is

Question

A projectile is thrown with a speed $u$ at an angle $	heta$ to the horizontal. The radius of curvature of its trajectory when the velocity vector of the projectile makes an angle $\alpha$ with the horizontal is

Dr. Abubacker · Accepted Answer

(c) Refer figure when the velocity vector makes an angle $\alpha$ with the horizontal, the component of acceleration, perpendicular to velocity, $ie,$ the centripetal acceleration is $g\cos\alpha$. As horizontal component of velocity remains unchanged in angular projection of projectile, hence
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$v\cos{\alpha = u\cos	heta}\ $or $v = \frac{u\cos	heta}{\cos\alpha}$
As, $g\cos\alpha$ provides centripetal acceleration, hence
$g\cos\alpha = \frac{v^{2}}{r}\ $ or $\frac{v^{2}}{g\cos\alpha} = \frac{u^{2}\cos^{2}	heta}{g\cos^{3}\alpha}$

Q. A projectile is thrown with a speed $u$ at an angle $\theta$ to the horizontal. The radius of curvature of its trajectory when the velocity vector of the projectile makes an angle $\alpha$ with the horizontal is

Explanation

Reviewed by Dr. Abubacker

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