---
title: "Projectile time of flight and range"
subject: "Applied Mechanics (Dynamics)"
chapter: 1
key: "semester_3/applied_mechanics_dynamics:projectile-slope-time-range"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/projectile-slope-time-range
site: BCT Engineering
---

# Projectile time of flight and range

Applied Mechanics (Dynamics) · Chapter 1: Curvilinear Motion of Particles

- Asked in: 2081 Bhadra, 3 marks
- Video: [A projectile is launched from point A with the initial conditions shown in the figure. Determine the](https://www.youtube.com/watch?v=f3XNYcvGMVI) — Dynamics Professor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/projectile-slope-time-range
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-1

## Question

A projectile is projected as shown in figure with an initial velocity of 30 m/s. Determine the time of flight and range. [Figure: launch point A on a slope of 3 vertical to 4 horizontal (angle $\theta$) at the launch end; the projectile lands at ground point B, 25 m below A.]

## Short answer

Resolve initial velocity along 3:4 slope: v_x = 24 m/s, v_y = 18 m/s. Vertical equation −25 = 18t − 4.905t² gives t = 4.74 s. Range = 24×4.74 = 113.9 m.

## Answer

**Assumption (figure not shown):** the slope at A is $3:4$, so $\tan\theta=3/4$, $\cos\theta=0.8$, $\sin\theta=0.6$. The 30 m/s launch is along the slope, upward. B is 25 m below A. Take $g=9.81\ \text{m/s}^2$ and neglect air resistance.

#### Components of launch velocity
$$
v_x=30(0.8)=24\ \text{m/s},\qquad v_y=30(0.6)=18\ \text{m/s}
$$

#### Time of flight
Vertical motion from A to B ($y=-25$ m):

$$
\begin{aligned}
-25&=18t-\tfrac12(9.81)t^{2}\\
4.905t^{2}-18t-25&=0\\
t&=\frac{18+\sqrt{18^{2}+4(4.905)(25)}}{2(4.905)}=4.744\ \text{s}
\end{aligned}
$$

#### Range
$$
R=v_x t=24(4.744)=113.86\ \text{m}
$$

**Answer: time of flight $t\approx 4.74$ s; horizontal range $R\approx 113.9$ m.**

## Check yourself

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

1. With tan(theta) = 3/4 and a launch speed of 30 m/s along the slope, what are the horizontal and vertical launch components?
   A. v_x = 18 m/s, v_y = 24 m/s
   B. v_x = 24 m/s, v_y = 18 m/s
   C. v_x = 30 m/s, v_y = 0
   D. v_x = 22.5 m/s, v_y = 40 m/s
   Answer: B. cos(theta) = 0.8 and sin(theta) = 0.6, so v_x = 24 and v_y = 18 m/s. (Key point: Launch components)

2. Which equation gives the time of flight from A to B (B is 25 m below A)?
   A. 25 = 18t - 4.905t^2
   B. -25 = 18t - 4.905t^2
   C. -25 = 24t - 4.905t^2
   D. -25 = -18t - 4.905t^2
   Answer: B. Taking up as positive, the displacement is -25 m, with initial vertical velocity +18 m/s. (Key point: Vertical equation for time)

3. Once t = 4.74 s is found, how is the range obtained?
   A. R = v_y t = 18(4.74)
   B. R = v_x t = 24(4.74)
   C. R = 30 t = 30(4.74)
   D. R = v_x t / 2
   Answer: B. Horizontal velocity is constant at 24 m/s (no air resistance), so R = 24(4.74) = 113.9 m. (Key point: Range)

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