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NEET PHYSICSEasy

The velocity of a projectile at the initial point A is (2i^+3j^) m/s(2\hat{i} + 3\hat{j}) \text{ m/s}. Its velocity (in m/s) at the point B (landing point on the same horizontal plane) 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. Decomposition of Velocity: The initial velocity vector is given as v⃗A=2i^+3j^\vec{v}_A = 2\hat{i} + 3\hat{j}. This means the horizontal component ux=2 m/su_x = 2 \text{ m/s} and the initial vertical component uy=3 m/su_y = 3 \text{ m/s}.
  2. Horizontal Motion: In projectile motion, the horizontal acceleration is zero (neglecting air resistance). Therefore, the horizontal component of velocity remains constant throughout the flight. At point B, vx=ux=2i^v_x = u_x = 2\hat{i} .
  3. Vertical Motion: The vertical component is affected by gravity. For a projectile landing on the same horizontal level as the projection point, the time of flight allows gravity to reverse the vertical velocity component exactly. The magnitude remains the same, but the direction is inverted. Thus, at point B, vy=−uy=−3j^v_y = -u_y = -3\hat{j} .
  4. Resultant Velocity: Combining the components, the velocity at point B is v⃗B=2i^−3j^\vec{v}_B = 2\hat{i} - 3\hat{j}.
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