JEE MAIN - Physics (2019 - 11th January Morning Slot - No. 3)

A body is projected at t = 0 with a velocity 10 ms^–1 at an angle of 60^o 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 :

2.8 m

5.1 m

2.5 m

10.3 m

Explanation

v_x = 10cos60^o = 5 m/s v_y = 10cos30^o = $$5\sqrt 3 $$ m/s

velocity after t = 1 sec.

v_x = 5 m/s

v_y = $$\left| {\left( {5\sqrt 3 - 10} \right)} \right|$$ m/s = 10 $$-$$ 5$$\sqrt 3 $$

a_n = $${{{v^2}} \over R} \Rightarrow \,R\,$$ = $${{v_x^2 + v_y^2} \over {{a_n}}}$$ = $${{25 + 100 + 75 - 100\sqrt 3 } \over {10\cos \theta }}$$

tan$$\theta $$ = $${{10 - 5\sqrt 3 } \over 5}$$ = 2 $$-$$ $${\sqrt 3 }$$ $$ \Rightarrow $$ $$\theta $$ = 15^o

R = $${{100\left( {2 - \sqrt 3 } \right)} \over {10\cos 15}} = 2.8m$$

JEE MAIN - Physics (2019 - 11th January Morning Slot - No. 3)

Explanation

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