Q. 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 R. Neglecting air resistance and taking acceleration due to gravity g = 10 $ms^{-2}$, the value of R is :

Solution:

$v_x = 10 \cos 60^{\circ} = 5 \; m/s$
$v_y = 10 \cos 30^{\circ} = 5 \sqrt{3} \; m/s$
velocity after t = 1 sec.
$v_x = 5 \; m/s$
$v_y = |( 5 \sqrt{3} - 10)| m/s = 10 - 5 \sqrt{3}$
$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} $
$ \tan\theta = \frac{10-5\sqrt{3}}{5} = 2 -\sqrt{3} \Rightarrow \theta =15^{\circ} $
$ R = \frac{100\left(2-\sqrt{3}\right)}{10\cos15} = 2.8 m $

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