---
title: "Rectilinear motion definitions; projectile at 244 m/s"
subject: "Applied Mechanics"
chapter: 8
key: "semester_1/applied_mechanics:uniform-rectilinear-motion-projectile-244"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/uniform-rectilinear-motion-projectile-244
site: BCT Engineering
---

# Rectilinear motion definitions; projectile at 244 m/s

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

- Asked in: 2073 Shrawan, 2+8 marks; 2069 Chaitra, 2+8 marks (asked 2 times)
- Video: [Dynamics - Lesson 1: Introduction and Constant Acceleration Equations](https://www.youtube.com/watch?v=7aMiZ3b0Ieg) — Jeff Hanson
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/uniform-rectilinear-motion-projectile-244
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-8

## Question

Define the uniformly rectilinear motion and the uniformly accelerated rectilinear motion. A projectile is fired with an initial velocity of 244 m/s at a target B located 610 m above the gun A and at a horizontal distance of 3658 m. Neglect air resistance, determine the value of the firing angle α.

## Short answer

Uniform rectilinear motion: constant velocity, zero acceleration. Uniformly accelerated rectilinear motion: constant acceleration; v=u+at, s=ut+½at², v²=u²+2as. For the projectile: using trajectory equation y = x tanα - gx²/(2v₀²cos²α) yields tanα = 0.5640 or 2.7542. Firing angle α = 29.4° (low) or 70.0° (high).

## Answer

#### Uniformly rectilinear motion
A particle moves along a straight line with **constant velocity**, so its acceleration is zero: v = constant, x = x₀ + vt. The x–t graph is a straight line and the v–t graph is horizontal.

#### Uniformly accelerated rectilinear motion
A particle moves along a straight line with **constant acceleration** a. Then:
- v = u + at
- s = ut + ½at²
- v² = u² + 2as
where u is the initial velocity, v the velocity after time t and s the displacement. The v–t graph is a straight inclined line and the x–t graph is a parabola. Free fall (a = g = 9.81 m/s²) is the common example.

#### Projectile problem
Take A as origin. Initial speed v₀ = 244 m/s, target at x = 3658 m, y = 610 m. The equation of the trajectory (no air resistance):

```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 × 3658² / (2 × 244²) = 1102.42
- Substituting: 610 = 3658 tan α − 1102.42(1 + tan²α)
- 1102.42 tan²α − 3658 tan α + 1712.42 = 0
- tan α = [3658 ± √(3658² − 4(1102.42)(1712.42))] / (2 × 1102.42) = (3658 ± 2414.5)/2204.84

So tan α = 0.5640 or 2.7542.

**Answer: α = 29.4° (low trajectory) or α = 70.0° (high trajectory).** Both angles hit B; the firing angle normally used is 29.4°.

## Check yourself

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

1. Which describes uniformly rectilinear motion?
   A. Straight line, constant acceleration
   B. Straight line, constant velocity with zero acceleration
   C. Curved path, constant speed
   D. Straight line with velocity proportional to time
   Answer: B. Uniform rectilinear motion has constant velocity (a = 0), so x = x0 + vt and the x-t graph is a straight line. (Key point: Uniform rectilinear motion)

2. For uniformly accelerated rectilinear motion, what do the v-t and x-t graphs look like?
   A. v-t horizontal, x-t straight
   B. v-t inclined straight line, x-t parabola
   C. Both are parabolas
   D. v-t parabola, x-t straight
   Answer: B. v = u + at is linear in t and s = ut + 1/2 at^2 is quadratic in t. (Key point: Graphs of accelerated motion)

3. Why does the projectile problem give two firing angles (29.4 and 70.0 deg)?
   A. The target can be hit by a low or a high trajectory because the quadratic in tan α has two roots
   B. Air resistance changes the angle
   C. Gravity acts twice
   D. Only one angle is valid; the other is an error
   Answer: A. Substituting into the trajectory equation gives a quadratic in tan α with two positive roots, both reaching B. (Key point: Two firing angles)

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