Question

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

Answer 1

Answer: 2.8

Horizontal component of velocity
$v_x=10cos60^{\circ }=5m/s$
vertical component of velocity
$v_y=10cos30^{\circ }=5\sqrt{3}m/s$
After $t=1$ sec.
Horizontal component of velocity $v_x=5m/s$
Vertical component of velocity
$v_y=|(5\sqrt{3}-10)|m/s=10-5\sqrt{3}$
Centripetal, acceleration $a_n=\frac{v^2}{R}$
$\Rightarrow R=\frac{v_x^2+v_y^2}{a_n}=\frac{25+100+75-100\sqrt{3}}{10cos\theta }...(i)$
From figure (using (i))
$tan\theta =\frac{10-5\sqrt{3}}{5}=2-\sqrt{3}$
$\Rightarrow \theta =15^{\circ }$
$R=\frac{100(2-\sqrt{3})}{10cos15}=2.8m$