---
title: "Projectile hitting target 600 m below"
subject: "Applied Mechanics"
chapter: 8
key: "semester_1/applied_mechanics:projectile-target-600m-below-slope-3-4"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/projectile-target-600m-below-slope-3-4
site: BCT Engineering
---

# Projectile hitting target 600 m below

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

- Asked in: 2068 Shrawan
- Video: [Physics 3.1: Projectile Motion on an Incline (4 of 7) Negative Slope](https://www.youtube.com/watch?v=FIhG0FlrxbE) — Michel van Biezen
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/projectile-target-600m-below-slope-3-4
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-8

## Question

A particle projected at an angle of θ to horizontal axis with an initial velocity of 61 m/sec hits a target located at 600 m below the horizontal axis and having the inclined slope of ¾ downward from the axis of projection to the target. Find the projected angle θ and the maximum height achieved by the particle from the target.

## Short answer

With slope 3 horizontal to 4 vertical, target is at x=450 m, y=-600 m. Solving trajectory equation gives projection angle θ ≈ 66.0°. Maximum height above the target is 758 m.

## Answer

#### Setting up
Origin at the point of projection, v₀ = 61 m/s, angle θ above the horizontal. The target lies 600 m below the horizontal axis on an incline. Trajectory:

y = x tan θ − g x²/(2v₀² cos²θ) = x tan θ − [g x²/(2v₀²)](1 + tan²θ)

#### Position of the target
The slope is quoted as 3/4.
- If the incline falls 3 m for every 4 m horizontally, the target is at x = 800 m. The least speed that can reach (800, −600) is √[g(√(800² + 600²) − 600)] = 62.6 m/s, which is more than 61 m/s. The discriminant is negative and no firing angle exists, so this reading cannot be the one intended.
- With the slope read as 3 horizontal to 4 vertical (the incline falls 4 m for every 3 m of horizontal distance), the target is at x = 600 × 3/4 = **450 m** and the problem has a solution. This is used below.

#### Firing angle
g x²/(2v₀²) = 9.81 × 450²/(2 × 61²) = 266.93

−600 = 450 tan θ − 266.93(1 + tan²θ)
266.93 tan²θ − 450 tan θ − 333.07 = 0
tan θ = [450 ± √(450² + 4 × 266.93 × 333.07)]/(2 × 266.93) = (450 ± 747.1)/533.87

tan θ = 2.2423 or −0.5565, so **θ = 65.96° above the horizontal** (the second root, θ = −29.1°, means firing below the horizontal).

#### Maximum height from the target
- Vertical component of velocity: v_y = 61 sin 65.96° = 55.7 m/s
- Rise above the axis: h = v_y²/(2g) = 55.7²/19.62 = 158.2 m
- Height above the target = 158.2 + 600 = **758.2 m**

**Answer: θ ≈ 66.0°; maximum height above the target ≈ 758 m.** (Time of flight = 18.1 s and 61 cos 65.96° × 18.1 = 450 m, which checks the target distance.)

## Check yourself

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

1. Why is the slope read as 3 horizontal to 4 vertical (x = 450 m)?
   A. The other reading (x = 800 m) needs at least 62.6 m/s, more than 61 m/s, so has no solution
   B. 450 is a round number
   C. It is shorter
   D. 800 gives a negative height
   Answer: A. With x = 800 the discriminant is negative. (Key point: Interpreting the slope)

2. Which equation is solved for tan θ?
   A. -600 = 450 tan θ - 266.93(1 + tan^2θ)
   B. 600 = 450 tan θ + 266.93
   C. -600 = 450 tan θ - 9.81
   D. 450 = 600 tan θ
   Answer: A. Substitute x = 450, y = -600 into the trajectory equation. (Key point: Trajectory equation)

3. Why is θ = 65.96 deg chosen?
   A. The other root (tan θ = -0.5565) is firing below the horizontal, but θ is the angle above it
   B. It is the smaller angle
   C. Both roots are above horizontal
   D. It is 45 deg
   Answer: A. The negative tan θ would fire downward. (Key point: Choosing the root)

4. How is maximum height above target obtained?
   A. v_y^2/2g only = 158.2 m
   B. Rise 158.2 m plus 600 m = 758.2 m
   C. 600 - 158.2
   D. v^2/2g = 189.7 m
   Answer: B. The peak is 158.2 m above the axis and the target is 600 m below it. (Key point: Maximum height from target)

## Close variants set in other exams

- [Projectile at 30°, 61 m/s to target below](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aprojectile-30deg-61ms-target-h-below) — Applied Mechanics, chapter 8, asked 2×

## Related in Applied Mechanics

- [Projectile target distance; dependent motion](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aprojectile-target-overshoot-undershoot-dependent-motion) — Applied Mechanics, chapter 8