A body is projected at $t=0$ with a velocity $10 \mathrm{~ms}^{-1}$ at an angle of $60^{0}$ with the horizontal. The radius of curvature of its trajectory at $t=1 \mathrm{~s}$ is R. Neglecting air resistance and taking acceleration due to gravity $g=10 \mathrm{~ms}^{-2}$, the radius of $R$ is:

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Kinematics

A body is projected at

t = 0

with a velocity

10 ms^{- 1}

at an angle of

6 0^{0}

with the horizontal. The radius of curvature of its trajectory at

t = 1 s

is R. Neglecting air resistance and taking acceleration due to gravity

g = 10 ms^{- 2}

, the radius of

R

is:

Answer: (B)

Resolve the launch velocity into

u_{x} = 10 cos 6 0^{\circ} = 5 m/s

and

u_{y} = 10 sin 6 0^{\circ} = 53 m/s

. The horizontal part stays

5 m/s

, while at

t = 1 s

the vertical part is

u_{y} - g t = 53 - 10 \approx - 1.34 m/s

, so the speed is

v = 5^{2} + 1.3 4^{2} \approx 5.18 m/s

. The radius of curvature is

R = \frac{v ^{2}}{a _{⊥}}

, where

a_{⊥}

is the component of gravity perpendicular to the velocity,

a_{⊥} = g \frac{v _{x}}{v} = 10 \times \frac{5}{5.18} \approx 9.66 m/s^{2}

. Hence

R = \frac{5.1 8 ^{2}}{9.66} \approx 2.8 m

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Single correct option