---
title: "Airplane dropping water on fire"
subject: "Applied Mechanics (Dynamics)"
chapter: 1
key: "semester_3/applied_mechanics_dynamics:airplane-water-drop-distance"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/airplane-water-drop-distance
site: BCT Engineering
---

# Airplane dropping water on fire

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

- Asked in: 2080 Baishakh, 4 marks
- Video: [Plane Dropping Water | Velocity and Acceleration (X & Y)](https://www.youtube.com/watch?v=WPYriE6OyB4) — Engineering Hack
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/airplane-water-drop-distance
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-1

## Question

An airplane used to drop water on brushfires is flying horizontally in a straight line at 315 km/h at an altitude of 80 m. Determine the distance $d$ at which the pilot should release the water so that it will hit the fire at B.

## Short answer

Convert speed: 315 km/h = 87.5 m/s. Time to fall 80 m: t = √(2×80/9.81) = 4.04 s. Release distance d = 87.5 × 4.04 ≈ 353 m.

## Answer

Once released, the water moves as a projectile with the plane's horizontal velocity and zero initial vertical velocity.

#### Data
$$
v_x=315\ \text{km/h}=\frac{315}{3.6}=87.5\ \text{m/s},\qquad h=80\ \text{m},\qquad g=9.81\ \text{m/s}^2
$$

#### Time to fall 80 m
$$
h=\tfrac12 g t^{2}\ \Rightarrow\ t=\sqrt{\frac{2(80)}{9.81}}=4.0386\ \text{s}
$$

#### Horizontal distance
$$
d=v_x t=87.5(4.0386)=353.4\ \text{m}
$$

**Answer: the water must be released $d\approx 353$ m before the fire (horizontal distance).**

## Check yourself

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

1. What are the water's initial velocity components just after release?
   A. Horizontal 87.5 m/s, vertical 0
   B. Horizontal 0, vertical 87.5 m/s downward
   C. Horizontal 315 m/s, vertical 0
   D. Horizontal 87.5 m/s, vertical 9.81 m/s
   Answer: A. The water keeps the plane's horizontal velocity and starts with zero vertical velocity. (Key point: Release conditions)

2. How is 315 km/h converted to m/s?
   A. Multiply by 3.6
   B. Divide by 3.6
   C. Divide by 60
   D. Multiply by 1000
   Answer: B. 1 km/h = 1/3.6 m/s, so 315/3.6 = 87.5 m/s. (Key point: Unit conversion)

3. Which relation gives the time to fall the 80 m?
   A. t = sqrt(2h/g)
   B. t = h/g
   C. t = sqrt(h/g)
   D. t = 2h/g
   Answer: A. From h = (1/2) g t^2, t = sqrt(2(80)/9.81) = 4.04 s. (Key point: Fall time)

4. How is the horizontal release distance d found?
   A. d = v_x t = 87.5(4.04)
   B. d = h
   C. d = (1/2) g t^2
   D. d = v_x / t
   Answer: A. Horizontal speed is constant, so d = 87.5 x 4.04 = about 353 m. (Key point: Release distance)

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