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A particle is projected on a planet with velocity vector V = 6 i-hat + (20 - 4t) j-hat m/s. Match the physical quantities in Column I with their correct values in Column II. Column I: (P) Time of flight (in seconds) (Q) Time (in seconds) when the particle moves at 53 degrees above the horizontal (upward direction) (R) Range of the particle (in meters) (S) Maximum height of the particle (in meters) Column II: (1) 3 (2) 10 (3) 50 (4) 60 (5) 70

1. P -> 2; Q -> 1; R -> 4; S -> 3
2. P -> 1; Q -> 2; R -> 3; S -> 4
3. P -> 2; Q -> 3; R -> 4; S -> 1
4. P -> 2; Q -> 4; R -> 1; S -> 3

Correct answer: P -> 2; Q -> 1; R -> 4; S -> 3

Solution

From the velocity: vx = 6 m/s (constant), vy(t) = 20 - 4t, so g = 4 m/s² on this planet. (P) Time of flight: T = 2*uy/g = 2*20/4 = 10 s -> value (2). (Q) At 53 deg above horizontal: tan(53 deg) = 4/3. vy/vx = 4/3 => (20-4t)/6 = 4/3 => 20-4t = 8 => t = 3 s -> value (1). (R) Range = vx * T = 6 * 10 = 60 m -> value (4). (S) H_max = uy²/(2g) = 20²/(2*4) = 400/8 = 50 m -> value (3). Matching: P->2, Q->1, R->4, S->3.

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