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Relative motion; projectile from A to B

Short answer

Relative motion is the motion of one particle as observed from another, e.g., two cars where B appears to move at 20 km/h east relative to A. For the projectile, using trajectory equation y = x tanα - gx²/(2v₀²cos²α) yields tanα = 0.7313 or 1.9871, giving α = 36.2° or 63.3°.

Explain about relative motion of particle with example. A projectile is fired from position A with an initial velocity of 200 m/sec at a target B on the right located 500 m above the position A. The horizontal distance between A and B is 3000 m. Determine the firing angle neglecting air resistance.

Relative motion of a particle

The motion of a particle B as seen from another moving particle A is its relative motion. With position vectors measured from a fixed origin:

  • r_B/A = r_B − r_A
  • v_B/A = v_B − v_A, so v_B = v_A + v_B/A
  • a_B/A = a_B − a_A, so a_B = a_A + a_B/A

Example: two cars A and B on a straight road moving east at 60 km/h and 80 km/h. The velocity of B relative to A is 80 − 60 = 20 km/h east: a passenger in A sees B pulling away at 20 km/h. Another example is a boat crossing a river, where the velocity of the boat relative to the ground is the vector sum of its velocity relative to water and the velocity of the water.

Projectile problem

Take A as origin: v₀ = 200 m/s, target B at x = 3000 m, y = +500 m.

x = (v₀ cos α) t
y = (v₀ sin α) t − ½ g t²
y = x tan α − g x² / (2 v₀² cos²α)
  = x tan α − [g x² / (2 v₀²)] (1 + tan²α)
  • g x²/(2v₀²) = 9.81 × 3000² / (2 × 200²) = 1103.63
  • 500 = 3000 tan α − 1103.63(1 + tan²α)
  • 1103.63 tan²α − 3000 tan α + 1603.63 = 0
  • tan α = [3000 ± √(3000² − 4(1103.63)(1603.63))]/(2 × 1103.63) = (3000 ± 1385.9)/2207.25

So tan α = 0.7313 or 1.9871.

Answer: α = 36.2° (low trajectory) or α = 63.3° (high trajectory).

✓ Answer checked by Claude Opus 5.5 (high effort)

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