A body is projected at t = 0 with a velocity 10 ms–1 at an angle of… | JEE Main 2019 (11 Jan, Shift 1)

Q47JEE Main 2019 · 11 JAN · SHIFT 1Difficult

Centripetal Acceleration & Force

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

$u_x = 10\cos 60^\circ = 5$ m/s, $u_y = 10\sin 60^\circ = 5\sqrt{3} \approx 8.66$ m/s.
At $t = 1$ s: $v_y = 8.66 - 10 = -1.34$ m/s, $v^2 = 25 + 1.80 = 26.8$ , $v \approx 5.18$ m/s.
Normal acceleration $a_n = g\cos\phi = g\frac{v_x}{v} = 10 \times \frac{5}{5.18} \approx 9.66$ m/s $^2$ .
$R = \frac{v^2}{a_n} = \frac{26.8}{9.66} \approx 2.8$ m
So the answer is (a) 2.8 m.

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