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EEJ MAIN Physics QUESTION #7092
Question 1
A body is projected at $10\text{ ms}^{-1}$ at $60^{\circ}$ to the horizontal. Calculate the radius of curvature at $t=1\text{ s}$ ($g=10\text{ ms}^{-2}$).
  • 10.3 m
  • 2.8 m✔️
  • 2.5 m
  • 5.1 m
Correct Answer Explanation

Radius of curvature $R = \frac{v^2}{a_{\perp}}$.

  • At $t=1$: $v_x = 10 \cos 60^{\circ} = 5$; $v_y = 10 \sin 60^{\circ} - 10(1) = 5\sqrt{3} - 10 \approx -1.34$.
  • Total $v = \sqrt{5^2 + (-1.34)^2} \approx 5.18$.
  • $a_{\perp} = g \cos \theta$ where $\theta$ is the angle of the velocity vector. After calculation, $R \approx 2.8\text{ m}$.