Question

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

• A. $10.3\text{m}$
• B. $2.8\text{m}$
• C. $2.5\text{m}$
• D. $5.1\text{m}$

Answer 1

The correct option is B $2.8\text{m}$


Horizontal component of velocity

$v_x=10\cos {60}^{\circ }=5\text{m/s}$

vertical component of velocity,

$v_y=10\cos {30}^{\circ }=5\sqrt{3}\text{m/s}$

After $t=1\text{sec}$

Horizontal component of velocity $v_{x1}=5\text{m/s}$

Vertical component of velocity,

$v_{1y}=|(5\sqrt{3}-10)|\text{m/s}=10-5\sqrt{3}$

Centripetal acceleration, $a_n=\frac{v^2}{R}$

$\Rightarrow R=\frac{v_x^2+v_y^2}{a_n}=\frac{25+100+75-100\sqrt{3}}{10\cos \theta }\dots \left(i\right)$

So from the given diagram,

$\tan \theta =\frac{10-5\sqrt{3}}{5}=2-\sqrt{3}$

So, $\cos \theta =\frac{1}{4(2-\sqrt{3})}$

Putting the value of $\cos \theta$ in eq. $\left(i\right)$,

$\therefore R=\frac{100(2-\sqrt{3})}{10\times \frac{1}{4(2-\sqrt{3})}}=2.8\text{m}$

Hence, $\left(B\right)$ is the correct answer.