---
title: "Projectile at 30°, 61 m/s to target below"
subject: "Applied Mechanics"
chapter: 8
key: "semester_1/applied_mechanics:projectile-30deg-61ms-target-h-below"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/projectile-30deg-61ms-target-h-below
site: BCT Engineering
---

# Projectile at 30°, 61 m/s to target below

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

- Asked in: 2067 Asar, 8 marks
- Video: [Projectile Motion: 3 methods to answer ALL questions!](https://www.youtube.com/watch?v=1eRimU5befo) — That Physics Guy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/projectile-30deg-61ms-target-h-below
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-8

## Question

A particle is projected at an angle of 30° to horizontal axis with an initial velocity of 61 m/sec hit the target located at 'h' meter below the horizontal axis and having the inclined slope of ¾ downward from the axis of the target. Find the sloping distance covered by the projectile and the maximum height achieved by particle from the target.

## Short answer

Trajectory equation y = x tan30° – gx²/(2v₀²cos²30°) set equal to –0.75x gives x = 755.2 m, drop h = 566.4 m. Sloping distance = √(x²+h²) = 944 m. Maximum height above projection is 47.4 m; height above target = 47.4 + 566.4 = 614 m.

## Answer

#### Setting up
v₀ = 61 m/s at θ = 30°. The target lies on an incline falling 3 m for every 4 m horizontally (tan φ = 3/4, φ = 36.87°). Origin at the point of projection; the target is at (x, −0.75x).

Trajectory: y = x tan θ − g x²/(2v₀² cos²θ)
- tan 30° = 0.5774
- g/(2v₀² cos²30°) = 9.81/(2 × 3721 × 0.75) = 0.0017575

#### Target position
−0.75x = 0.5774x − 0.0017575x²
0.0017575x = 0.5774 + 0.75 = 1.3274, so **x = 755.2 m**
h = 0.75 × 755.2 = **566.4 m** below the axis.

#### Sloping distance
Sloping distance = √(755.2² + 566.4²) = 755.2/cos 36.87° = **944.0 m**
(Time of flight: t = 755.2/(61 cos 30°) = 14.30 s.)

#### Maximum height from the target
- Vertical velocity: v_y = 61 sin 30° = 30.5 m/s
- Rise above the axis: 30.5²/(2 × 9.81) = 47.41 m
- Height above the target: 47.41 + 566.4 = **613.8 m**

**Answer: sloping distance ≈ 944 m; maximum height above the target ≈ 614 m.**

## Check yourself

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

1. How is the target position expressed?
   A. On the line y = -0.75x (slope 3 down per 4 across)
   B. y = -x
   C. y = 0.75x
   D. x = 0.75y
   Answer: A. Slope falls 3 m per 4 m horizontally. (Key point: Target line)

2. What is x?
   A. 944 m
   B. 566.4 m
   C. 755.2 m
   D. 613.8 m
   Answer: C. -0.75x = 0.5774x - 0.0017575x^2 gives x = 1.3274/0.0017575 = 755.2 m. (Key point: Solving for x)

3. How is sloping distance found?
   A. sqrt(755.2^2 + 566.4^2) = 944 m
   B. 755.2 m
   C. 566.4 m
   D. 755.2 + 566.4
   Answer: A. Hypotenuse of the horizontal and vertical offsets. (Key point: Sloping distance)

4. What is the max height above target?
   A. 47.41 m
   B. 566.4 m
   C. 613.8 m (47.41 + 566.4)
   D. 755 m
   Answer: C. Rise above the axis (30.5^2/2g) plus the depth of the target. (Key point: Maximum height from target)

## Close variants set in other exams

- [Projectile hitting target 600 m below](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aprojectile-target-600m-below-slope-3-4) — 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
- [Relative motion; projectile from A to B](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Arelative-motion-projectile-a-b) — Applied Mechanics, chapter 8
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- [Rectilinear motion definitions; projectile at 244 m/s](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Auniform-rectilinear-motion-projectile-244) — Applied Mechanics, chapter 8, asked 2×

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