Question

A particle is projected horizontally from the top of a tower with a velocity $v_0$. If $v$ be its velocity at any instant, then the radius of curvature of the path of the particle at that instant is directly proportional to

• A. $v^3$
• B. $v^2$
• C. $v$
• D. $\frac{1}{v}$

Answer 1

The correct option is B $v^2$

$mg=m{v}^2/r$

$r_t=v^2/g$

Hence, it is proportional to $v^2$