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

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A

$2.8m$

B

$2.5m$

C

$10.3m$

D

$5.1m$

Solution:

At $t=1s$ $V_{y}=5\sqrt{3}-10\times 1$
$V=\sqrt{25 + \left(\right. 5 \sqrt{3} - 10 \left(\left.\right)^{2}}$
$V_{x}=5$
$=\sqrt{25 + 75 + 100 - 100 \sqrt{3}}=\sqrt{200 - 100 \sqrt{3}}=5.11$
$r_{c}=\frac{v^{2}}{g cos \phi}=\frac{v^{2}}{g \frac{u_{x}}{v}}=\frac{v^{3}}{g u_{x}}=\frac{\left(\right. 5 . 11 \left(\left.\right)^{3}}{10 \times 5}=2.77m$

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