A body is projected at t = 0 with a velocity 10 ms–1 at an angle of 600 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:

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2.5 m

10.3 m

2.8 m

5.1 m

answer is C.

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Detailed Solution

vx=10cos600=5 m/s vy=10cos300=53 m/sVelocity after t = 1 sec. vx=5 m/svy=|(53−10)|m/s=10−53an=v2R⇒R=vx2+vy2an=25+100+75−100310cosθtanθ=10−535=2−3⇒θ=150 R=100(2−3)10cos15=2.8m

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