Question

A body is projected at$t=0$with a velocity $10m{s}^{–1}$ at an angle of $60°$ 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. $2.5m$
• B. $10.3m$
• C. $2.8m$
• D. $5.1m$

Answer 1

The correct option is C

$2.8m$

Step 1. Given Data:

Initial velocity, $u=10m/s$

Angle of projection, $\theta ={60}^o$

Radius of curvature at $t=1s$ is $R$

Acceleration due to gravity, $g=10m/s$

Step 2. Finding the radius of curvature:

Resolving the velocity into $x$ and $y$ components, at any time $t$

$v_x=u\cos {60}^o=10\times \frac{1}{2}$

$v_x=5m/s$

$v_y=u\sin {60}^o+at=\frac{10\sqrt{3}}{2}-10t$

$v_y=\left|5\sqrt{3}-10t\right|$

At $t=1s$

$v_y=\left|5\sqrt{3}-10\right|$

Angle made by body with respect to horizontal axis at $t=1s$

$\tan \alpha =\left|\frac{v_y}{v_x}\right|=\left|\frac{5\sqrt{3}-10}{5}\right|$,

$\tan \alpha ==\left|\sqrt{3}-2\right|$

$\alpha ={\tan }^{-1}\left(\left|\sqrt{3}-2\right|\right)={15}^o$

The radius of curvature same as range at time $t$, which is given by

$R=\frac{v^2}{g\cos \alpha }=\frac{{v_x}^2+{v_y}^2}{g\cos \alpha }$

$R=\frac{5^2+{\left(5\sqrt{3}-10\right)}^2}{10\times \cos {15}^o}$

$R=\frac{25+75+100-100\sqrt{3}}{10\times \cos {15}^o}$

$R=\frac{26.79}{9.65}=2.77$

$R=2.77m\approx 2.8m$

Hence, option C is correct