Question

A body is projected at $t = 0$ with a velocity $10 \,ms^{-1}$ 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 $g = 10\, ms^{-2}$. the value of $R$ is:

Answer 1

Horizontal component of velocity
$v_x = 10\,cos \,60^{\circ} = 5 \,m/s$
vertical component of velocity
$v_y = 10\, cos \,30^{\circ} = 5\sqrt{3} \,m/s$
After $t = 1 $ sec.
Horizontal component of velocity $v_x = 5 \,m/s$
Vertical component of velocity
$v_y = |(5\sqrt{3} - 10)| m/s = 10 - 5 \sqrt{3}$
Centripetal, acceleration $a_n = \frac{v^2}{R}$
$\Rightarrow R = \frac{v^2_x + v^2_y}{a_n} = \frac{25 + 100 + 75 - 100\sqrt{3}}{10\,cos\,\theta} ...(i)$
From figure (using (i))
$tan\,\theta = \frac{10 - 5\sqrt{3}}{5} = 2 - \sqrt{3}$
$\Rightarrow \theta = 15^{\circ}$
$R = \frac{100( 2 - \sqrt{3})}{10\,cos \,15} = 2.8\,m$