Question

A body is projected at $t=0$ with a velocity $10{\text{ms}}^{-1}$ at an angle of ${60}^{\circ }$ with the horizontal. The radius of curvature of its trajectory at $t=1\text{s}$ is $R$. Neglecting air resistance and taking acceleration due to gravity $g=10{\text{ms}}^{-2}$, the value of $R$ is:
$(\text{Take:}\tan {15}^{\circ }=0.268;\cos {15}^{\circ }=0.966)$

• 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 $2.8\text{m}$


At $t=0$,
Horizontal component of velocity
$u_x=10\cos {60}^{\circ }=5{\text{ms}}^{-1}$

Vertical component of velocity
$u_y=10\cos {30}^{\circ }=5\sqrt{3}{\text{ms}}^{-1}$

At $t=1\text{s}$,
Horizontal component of velocity
$v_x=5{\text{ms}}^{-1}$

Vertical component of velocity
$v_y=u_y-gt=|(5\sqrt{3}-10)|{\text{ms}}^{-1}$

$\Rightarrow v_y=10-5\sqrt{3}{\text{ms}}^{-1}$

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

$\Rightarrow R=\frac{v_x^2+v_y^2}{a_c}$

Here, $g\cos \theta$ points towards the centre and acts as the centripetal acceleration,

$\Rightarrow R=\frac{5^2+(10-5\sqrt{3}{)}^2}{g\cos \theta }$

$\Rightarrow R=\frac{25+(100+75-100\sqrt{3})}{10\cos \theta }$

$\Rightarrow R=\frac{100(2-\sqrt{3})}{10\cos \theta }$

At $t=1\to \frac{v_y}{v_x}=\tan \theta$,

$\tan \theta =\frac{10-5\sqrt{3}}{5}=2-\sqrt{3}=0.268$

$\Rightarrow \theta ={15}^{\circ }$

$\Rightarrow R=\frac{100(2-\sqrt{3})}{10\cos 15}=\frac{100\times 0.268}{10\times 0.966}=2.8\text{m}$

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