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Physics · MOTION IN A PLANE

हिन्दी

In the given parabolic path of a projectile shown in the figure, at the highest point of its trajectory, what are the vertical component of velocity (\( v_y \)) and the angle (\( \theta \)) made by the net velocity vector with the horizontal?

  1. \( v_y = 0 \) and \( \theta = 0^\circ \)
  2. \( v_y = v_o\sin\theta_o \) and \( \theta = 90^\circ \)
  3. \( v_y = v_o\cos\theta_o \) and \( \theta = 45^\circ \)
  4. \( v_y = -v_o\sin\theta_o \) and \( \theta = 180^\circ \)

Correct answer · Option 1

\( v_y = 0 \) and \( \theta = 0^\circ \)

Explanation

At the peak of the parabolic path, the vertical velocity becomes zero (\( v_y = 0 \)) while horizontal velocity \( v_x = v_o\cos\theta_o \) remains. Hence, net velocity is purely horizontal, making \( \theta = 0^\circ \).
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