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

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$10.3 m$

$2.8 m$

$2.5 m$

$5.1 m$

answer is A.

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Detailed Solution

At $t = 1 s$

$V_{y} = 5 \sqrt{3} - 10 \times 1$

$V = \sqrt{25 + {(5 \sqrt{3} - 10)}^{2}}$

$V_{x} = 5$

$= \sqrt{25 + 75 + 100 - 100 \sqrt{3}} = \sqrt{200 - 100 \sqrt{3}} = 5.11$

$r_{c} = \frac{v^{2}}{g \cos ϕ} = \frac{v^{2}}{g \frac{u_{x}}{v}} = \frac{v^{3}}{g u_{x}} = \frac{{(5.11)}^{3}}{10 \times 5} = 2.77 m$

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