Question

A stone is thrown horizontally under gravity with a speed of $10m/\sec$ . Find the radius of curvature of its trajectory at the end of $3\sec$ after motion began.
(A) $10\sqrt {10} m$
(B) $100\sqrt {10} m$
(C) $\sqrt {10} m$
(D) $100m$

Answer 1

At the highest point, the velocity and its acceleration are perpendiculars. Hence, we can say that the projectile is in a circular motion like it is in the case of a body in a uniform circular motion. Thus the acceleration of the projectile at the point is $a = \dfrac{{{v^2}}}{R}$ , where $R$ is the radius of curvature.
Formula used:
$a = \dfrac{{{v^2}}}{R}$

Complete answer:
The radius of curvature of a path at a point in a circle to which the curve of the path touches the circle tangentially. It states exactly how much the curve is at this point. The less the radius of curvature, the more pointed is the curve at the given point. If the radius of curvature is infinite now it means that the curve is a straight line.
Let us calculate the radius of curvature $R$ for a projectile at its height point from the ground.
It is already given that the object is projected horizontally with a speed ${v_x} = 10m/s$
Now after $3\sec$ it will gain vertical speed so,
${v_y} = gt = 3 \times 10 = 30m/s$
Currently, we see that the angle that it will create with the horizontal is given as
$\tan \theta = \dfrac{{{v_y}}}{{{v_x}}}$
$\Rightarrow \theta = {71.56^ \circ }$
Here, we can say that the projectile is in a circular motion since the velocity and its acceleration are perpendicular like it is the case of a body in a uniform circular motion. And the radius of that circle will be the radius of curvature at the highest point.
At this time the acceleration perpendicular to its net speed is specified as,
$a = g\cos \theta$
$\Rightarrow 10\cos \left( {{{71.56}^ \circ }} \right)$
The relation between acceleration $a$ (centripetal acceleration) and velocity $v$ of the body in a uniform circular motion is $a = \dfrac{{{v^2}}}{R}$
In this case, $a = 10\cos \left( {{{71.56}^ \circ }} \right)$ and $v = \sqrt {{v_x}^2 + {v_y}^2}$ . substitute these values in the above equation.
$10\cos \left( {{{71.56}^\circ }} \right) = \dfrac{{{v_x} + {v_y}}}{R}$
Taking the $R$ in the left-hand side we get,
$R = \dfrac{{{{10}^2} + {{30}^2}}}{{10\cos \left( {{{71.56}^ \circ }} \right)}}$
On further solving we get,
$R = 100\sqrt {10}$
Therefore, the radius of curvature of the trajectory of the stone at the end of $3\sec$ after motion began is $100\sqrt {10} m$ .

Hence, the correct answer is (B) $100\sqrt {10} m$ .

Note:
In a circular motion we know that there should be a force that is responsible for the circular motion of the particle or body. That particular force is known as the centripetal force. The direction of the acceleration is also towards the center of the trajectory. So, in a circular motion, the direction of the centripetal force and the direction of the centripetal acceleration is in the same direction.