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NEET PHYSICSEasy

The velocity of a projectile at the initial point AA is (2i^+3j^) m/s(2\hat{i} + 3\hat{j}) \text{ m/s}. Its velocity (in m/s) at point BB (the point where it hits the ground at the same horizontal level) is:

A

−2i^+3j^-2\hat{i} + 3\hat{j}

B

−2i^−3j^-2\hat{i} - 3\hat{j}

C

2i^−3j^2\hat{i} - 3\hat{j}

D

2i^+3j^2\hat{i} + 3\hat{j}

Step-by-Step Solution

  1. Concept: In projectile motion (neglecting air resistance), the acceleration due to gravity acts only in the vertical direction (downward). There is no acceleration in the horizontal direction.
  2. Horizontal Component: Since ax=0a_x = 0, the horizontal component of velocity remains constant throughout the flight. Given vix=2 m/sv_{ix} = 2 \text{ m/s}, then vfx=2 m/sv_{fx} = 2 \text{ m/s}.
  3. Vertical Component: Gravity accelerates the particle downwards. At the same vertical level (launch and landing points), the magnitude of the vertical velocity is the same, but the direction is reversed (upward at A, downward at B). Given viy=3 m/sv_{iy} = 3 \text{ m/s} (upward), then vfy=−3 m/sv_{fy} = -3 \text{ m/s} (downward).
  4. Conclusion: The velocity vector at point B is the vector sum of these components: v⃗B=vfxi^+vfyj^=2i^−3j^\vec{v}_B = v_{fx}\hat{i} + v_{fy}\hat{j} = 2\hat{i} - 3\hat{j}
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