Question

A stone is thrown horizontally with a velocity of $10m/sec$. Find the radius of curvature of it's 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

The correct option is B $100\sqrt{10}m$

Given,

$v_x=10m/s$

$t=3s$

Velocity along y -axis after $t=3s$,

$v_y=gt=10\times 3$

$v_y=30m/s$

Velocity of stone after $t=3sec$ is,

$\overrightarrow{v}=v_x\hat{i}+v_y\hat{j}$

$\overrightarrow{v}=(10\hat{i}+30\hat{j})m/s$

$tan\theta =\frac{v_y}{v_x}$

$tan\theta =\frac{30}{10}=3$

$\theta =ta{n}^{-1}\left(3\right)={71.56}^0$

The acceleration perpendicular to its net speed is,

$a=gcos\theta$

$a=10cos\left({71.56}^0\right)$

Radius of curvature, $r=\frac{v^2}{a}$

$r=\frac{(10{)}^2+(30{)}^2}{10cos{71.56}^0}$

$r=100\sqrt{10}m$

The correct option is B.