---
title: "Relative motion; projectile from A to B"
subject: "Applied Mechanics"
chapter: 8
key: "semester_1/applied_mechanics:relative-motion-projectile-a-b"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/relative-motion-projectile-a-b
site: BCT Engineering
---

# Relative motion; projectile from A to B

Applied Mechanics · Chapter 8: Kinematics of Particles and Rigid Body

- Asked in: 2074 Asoj, 2+8 marks
- Video: [Relative Motion Analysis of Two Particles Using Translating Axes (learn to solve any problem)](https://www.youtube.com/watch?v=opVKNCedkRo) — Question Solutions
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/relative-motion-projectile-a-b
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-8

## Question

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.

## 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°.

## Answer

#### 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.

```text
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).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the velocity of B relative to A?
   A. v_A - v_B
   B. v_B + v_A
   C. v_B - v_A
   D. v_B x v_A
   Answer: C. v_B/A = v_B - v_A, so v_B = v_A + v_B/A. (Key point: Relative velocity formula)

2. Cars A and B go east at 60 and 80 km/h. What does a passenger in A see?
   A. B approaching at 140 km/h
   B. B pulling away at 20 km/h east
   C. B at rest
   D. B moving west at 20 km/h
   Answer: B. v_B/A = 80 - 60 = 20 km/h east, so B pulls away. (Key point: Cars example)

3. How is the firing angle found for target B at x = 3000 m, y = 500 m?
   A. Use R = v^2 sin2α/g directly
   B. Substitute x and y into y = x tan α - [gx^2/(2v0^2)](1 + tan^2α) and solve the quadratic in tan α
   C. Assume α = 45 deg
   D. Use v = u + at only
   Answer: B. The target is not at launch level, so the range formula does not apply; the trajectory equation gives a quadratic with roots tan α = 0.7313 and 1.9871. (Key point: Trajectory method)

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