Question

A particle is projected horizontally from the top of the cliff of height h with a velocity of $\sqrt{2gh}$. The radius of curvature of the trajectory at the instant of projection will be?

• A. H/2
• B. H
• C. 2H
• D. $\infty$

Answer 1

Answer: (C)

2H

Speed of projection $v=\sqrt{2gH}$

The acceleration of the body at this moment $=g$

Assuming the motion to be circular momentartily, the acceleration of the body would be given as,

$g=\frac{v^2}{R}$

$R=\frac{v^2}{g}$

$R=\frac{2gH}{g}$

$R=2H$