Question:
A body is projected at t = 0 with a velocity 10 ms\(^{-1}\) 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 = 10 ms\(^{-2}\), the value of R is :
A body is projected at t = 0 with a velocity 10 ms\(^{-1}\) 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 = 10 ms\(^{-2}\), the value of R is :
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Radius of curvature \(R = v^2/a_\perp\), where \(a_\perp\) is the component of g perpendicular to the velocity at t = 1 s.
Updated On: Nov 5, 2026
- 2.8 m
- 5.1 m
- 2.5 m
- 10.3 m
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The Correct Option is A
Solution and Explanation
Components at t = 1 s: \(v_x = 10\cos60^\circ = 5\) m/s, \(v_y = 10\sin60^\circ - gt = 8.66 - 10 = -1.34\) m/s.
Speed: \(v^2 = 25 + 1.80 = 26.8\), so \(v = 5.18\) m/s.
Angle of velocity with horizontal: \(\cos\theta = v_x/v = 5/5.18 = 0.966\).
The component of g perpendicular to the velocity is \(a_\perp = g\cos\theta = 9.66\) m/s\(^2\).
\(R = \frac{v^2}{a_\perp} = \frac{26.8}{9.66} = 2.77\) m \(\approx 2.8\) m.
Answer: (1) 2.8 m
Speed: \(v^2 = 25 + 1.80 = 26.8\), so \(v = 5.18\) m/s.
Angle of velocity with horizontal: \(\cos\theta = v_x/v = 5/5.18 = 0.966\).
The component of g perpendicular to the velocity is \(a_\perp = g\cos\theta = 9.66\) m/s\(^2\).
\(R = \frac{v^2}{a_\perp} = \frac{26.8}{9.66} = 2.77\) m \(\approx 2.8\) m.
Answer: (1) 2.8 m
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