Question

A body is projected at $t = 0$ with a velocity of $10m{s^{ - 1}}$ at an angle of ${60^ \circ }$ with the horizontal. The radius of curvature of its trajectory at $t = 1s$ is $R$. Neglecting air resistance and taking acceleration due to gravity $g = 10m{s^{ - 2}}$, the value of $R$ is:
(A) $2.4m$
(B) $10.3m$
(C) $2.8m$
(D) $5.1m$

Answer 1

Hint
For solving this question, we need to use the concept of centripetal acceleration and find its value. From there, we can easily find the radius of curvature of the trajectory.
Formula Used: The formulae used in solving this question are:
-$v = u + at$, $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration, and $t$ is the time elapsed.
-${a_c} = \dfrac{{{v^2}}}{R}$, ${a_c}$ is the centripetal acceleration, $v$ is the velocity, and $R$ is the radius of the circle.

Complete step by step answer
We begin by finding the velocity of the body at $t = 1s$.
According to the question the initial velocity of the body, $u = 10m{s^{ - 1}}$
Also, the angle of projection $\theta = {60^ \circ }$
Horizontal direction:
Let ${u_h}$and${v_h}$ be the horizontal component of the initial and the final velocity of the body respectively.
$\Rightarrow {u_h} = u\cos \theta $
$\Rightarrow {u_h} = 10\cos {60^ \circ }$
On solving, we get
$\Rightarrow {u_h} = 10(0.5) = 50m{s^{ - 1}}$
The acceleration in the horizontal direction is zero as there is no horizontal force on the body, i.e.
$\Rightarrow {a_h} = 0$
By the first equation of motion
$\Rightarrow {v_h} = {u_h} + {a_h}t$
$\Rightarrow {v_h} = 5 + (0)1$
On solving, we get
$\Rightarrow {v_h} = 5m{s^{ - 1}}$ (1)
Vertical direction:
Let ${u_v}$ and${v_v}$ be the horizontal component of the initial and the final velocity of the body respectively.
$\Rightarrow {u_v} = u\sin \theta $
$\Rightarrow {u_v} = 10\sin {60^ \circ }$
On solving, we get
$\Rightarrow {u_v} = 10\dfrac{{\sqrt 3 }}{2} = 5\sqrt 3 m{s^{ - 1}}$
As the only force which acts on the body is the gravitational force, which is downwards, so the acceleration of the body is in the vertical direction, i.e.
$\Rightarrow {a_v} = g = - 10m{s^{ - 2}}$
By the first equation of motion, we have
$\Rightarrow {v_v} = {u_v} + {a_v}t$
$\Rightarrow {v_v} = 5\sqrt 3 + ( - 10)1$
On solving, we get
$\Rightarrow {v_v} = 5\left( {\sqrt 3 - 2} \right)m{s^{ - 1}}$ (2)
Now, the total final velocity $v$ of the body is given by
$\Rightarrow {v^2} = {v_h}^2 + {v_v}^2$
Substituting from (1) and (2), we get
$\Rightarrow {v^2} = {5^2} + {\left[ {5\left( {\sqrt 3 - 2} \right)} \right]^2}$
$\Rightarrow {v^2} = 26.79{m^2}{s^{ - 2}}$ (3)
Also, the angle $\varphi $ with the vertical made by the final velocity vector is given by
$\Rightarrow \tan \varphi = \dfrac{{{v_h}}}{{{v_v}}}$
$\Rightarrow \tan \varphi = \dfrac{5}{{5\left( {\sqrt 3 - 2} \right)}}$
On solving, we get
$\Rightarrow \tan \varphi = \dfrac{1}{{\left( {\sqrt 3 - 2} \right)}}$
Taking tangent inverse both the sides, we get
$\Rightarrow \varphi = {\tan ^{ - 1}}\left( {\dfrac{1}{{\left( {\sqrt 3 - 2} \right)}}} \right)$
$\Rightarrow \varphi = - {75^ \circ }$
Now, we find the component of the acceleration perpendicular to the velocity, that is, the centripetal acceleration of the body.
As the acceleration is vertically downwards, the centripetal acceleration is given by
$\Rightarrow {a_c} = {a_v}\sin \varphi $
$\Rightarrow {a_c} = - 10\sin \left( { - {{75}^ \circ }} \right)$
On solving, we get
$\Rightarrow {a_c} = 9.66m{s^{ - 2}}$ (4)
We know that the centripetal acceleration is given by
$\Rightarrow {a_c} = \dfrac{{{v^2}}}{R}$
Substituting from (3) and (4), we get
$\Rightarrow 9.66 = \dfrac{{26.79}}{R}$
By cross multiplication, we get
$\Rightarrow R = \dfrac{{26.79}}{{9.66}}$
$\Rightarrow R = 2.77m$
$\Rightarrow R \approx 2.8m$
So the radius of curvature of its trajectory at $t = 1s$is$2.8m$
Hence, the correct answer is option (C), $2.8m$

Note
Do not forget to find out the perpendicular component of the acceleration along the velocity of the body. As we know that the centripetal force acts perpendicular to the motion of the body, so the centripetal acceleration is the component of the acceleration perpendicular to the body.