---
title: "Projectile from hill, height, speed, time"
subject: "Applied Mechanics (Dynamics)"
chapter: 1
key: "semester_3/applied_mechanics_dynamics:bullet-hill-projectile"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/bullet-hill-projectile
site: BCT Engineering
---

# Projectile from hill, height, speed, time

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

- Asked in: 2072 Chaitra, 6 marks
- Video: [Projectile launched off a cliff at an angle](https://www.youtube.com/watch?v=TQSvfrSJucI) — Physics Ninja
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/bullet-hill-projectile
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-1

## Question

A bullet is fired at an angle of $30^\circ$ to the horizontal from a point 'P' on a hill and it strikes a target which is 100 m lower than the level of projection. The initial velocity of the bullet is 120 m/s. Neglecting the air resistance calculate: i) The maximum height to which the bullet will rise above the horizontal ii) The actual velocity with which it will strike the target iii) The total time required for the flight of bullet

## Short answer

Resolve initial velocity: v_x = 103.9 m/s, v_y0 = 60 m/s. Maximum height above projection is v_y0²/(2g) = 183.5 m. Final speed from v_y² = v_y0² + 2g(100 m) gives 127.9 m/s at 35.7° below horizontal; time of flight 13.72 s.

## Answer

Take $g=9.81\ \text{m/s}^2$ and the origin at P, with $y$ upward.

#### Launch components
$$
v_x=120\cos30^\circ=103.92\ \text{m/s},\qquad v_{y0}=120\sin30^\circ=60\ \text{m/s}
$$

#### i) Maximum height above the level of projection
At the top $v_y=0$:

$$
H=\frac{v_{y0}^{2}}{2g}=\frac{60^{2}}{2(9.81)}=183.49\ \text{m}
$$

#### ii) Velocity on striking the target ($y=-100$ m)
$$
v_y^{2}=v_{y0}^{2}+2g(100)=3600+1962=5562\ \Rightarrow\ v_y=74.58\ \text{m/s (downward)}
$$

$$
v=\sqrt{103.92^{2}+74.58^{2}}=127.9\ \text{m/s}
$$

Direction: $\tan^{-1}(74.58/103.92)=35.7^\circ$ below the horizontal.

#### iii) Total time of flight
$$
t=\frac{v_{y0}+|v_y|}{g}=\frac{60+74.58}{9.81}=13.72\ \text{s}
$$

**Answer: (i) $H=183.5$ m; (ii) $v=127.9$ m/s at $35.7^\circ$ below horizontal; (iii) $t=13.72$ s.**

## Check yourself

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

1. What are the launch components for 120 m/s at 30 degrees?
   A. v_x = 60, v_y = 103.9 m/s
   B. v_x = 103.9, v_y = 60 m/s
   C. v_x = 120, v_y = 0
   D. v_x = 83.1, v_y = 83.1 m/s
   Answer: B. v_x = 120 cos30 = 103.9; v_y0 = 120 sin30 = 60 m/s. (Key point: Launch components)

2. How is the maximum height above the level of projection found?
   A. H = v_y0^2 / (2g) = 183.5 m
   B. H = v_y0^2 / g
   C. H = 183.5 + 100
   D. H = v_x^2 / (2g)
   Answer: A. At the top v_y = 0. The 100 m depth of the target does not affect the height above P. (Key point: Maximum height)

3. How is the downward vertical speed at the target found?
   A. v_y^2 = 60^2 + 2g(100) = 5562, so v_y = 74.6 m/s
   B. v_y^2 = 60^2 - 2g(100)
   C. v_y = 60 m/s
   D. v_y = 100 / 13.72
   Answer: A. The bullet falls 100 m below P, so gravity does positive work of 2g(100). (Key point: Impact vertical speed)

4. How is the total flight time found?
   A. t = 60/g
   B. t = (v_y0 + |v_y|)/g = (60 + 74.58)/9.81 = 13.72 s
   C. t = 100/v_x
   D. t = 2(60)/g
   Answer: B. The vertical velocity changes from +60 to -74.58 m/s at rate g. (Key point: Time of flight)

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