2.2 Model projectile motion under uniform gravity | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Curvilinear and Relative Motion

Separating horizontal and vertical motion to predict a particle’s path

A projectile is an idealized particle that moves through the air while gravity is the only force acting on it. This model neglects air resistance and treats gravitational acceleration as constant. It helps predict where a launched object will land, how long it will be in flight, and how its velocity changes. Use one coordinate for horizontal motion and another for vertical motion: gravity changes vertical velocity, but not horizontal velocity. We take the observer to be stationary relative to the ground and use a ground-fixed frame. Unless stated otherwise, positive

x

points right and positive

y

points upward.

What you will learn

Define the particle, observer, reference frame, coordinates, and sign convention for a projectile problem.
Use uniform gravitational acceleration to model horizontal and vertical motion independently.
Determine a projectile’s position, velocity, flight time, peak height, or horizontal range from its initial conditions.
Check results using units, directions, initial conditions, and limiting cases.

1. Define the system and model

The system is the launched object, represented as a particle: its size and rotation are not part of the model. The observer measures its motion in a ground-fixed reference frame. The particle’s position is described by coordinates

x

and

y

, measured from a chosen origin. Displacement is the change in position; the path is the curve followed; and path length is the total distance travelled along that curve. These quantities are not interchangeable.

After release, assume the only force is the object’s weight, directed downward. Newton’s second law then gives constant acceleration: zero horizontally and

g

downward vertically, where

g=9.81\ \mathrm{m/s^2}

. This is the uniform-gravity model. The object’s mass does not affect its resulting acceleration because mass cancels when weight is used in Newton’s second law.

The initial velocity is often given by a speed and launch angle. If the angle

\theta

is measured above the positive horizontal direction, its components are

v_{0x}=v_0\cos\theta

and

v_{0y}=v_0\sin\theta

. The signs follow the chosen axes. A launch below horizontal has a negative vertical component.

\mathbf{a}=0\,\mathbf{i}-g\,\mathbf{j}

System: the projectile, modeled as a particle.
Observer and frame: a stationary observer in a ground-fixed frame.
Only force after release: weight downward; air resistance is neglected.
Choose and state the origin and positive coordinate directions.

2. Governing equations and how to use them

Because acceleration is constant, apply the constant-acceleration relations separately along each axis. Horizontally, acceleration is zero, so horizontal velocity stays constant. Vertically, acceleration is

-g

, so vertical velocity decreases as the projectile rises and becomes more negative as it falls.

For initial position

(x_0,y_0)

and initial velocity components

(v_{0x},v_{0y})

, the coordinates at time

t

follow from the position equations. The velocity components follow from the same constant-acceleration model. These equations apply from

t=0

until the event of interest, such as landing.

Choose the unknown that best matches the question. If time is known, find position or velocity directly. If the landing height is known, solve the vertical position equation for time, then use that time in the horizontal equation. At the highest point, vertical velocity is zero. Reject a negative time root if the requested event occurs after launch.

x=x_0+v_{0x}t,\quad y=y_0+v_{0y}t-\frac{1}{2}gt^2,\quad v_x=v_{0x},\quad v_y=v_{0y}-gt

Horizontal and vertical equations describe components of one motion.
Time is shared by both components; it links horizontal travel to vertical fall.
At the highest point, vertical velocity is zero, but horizontal velocity is generally not.

3. Useful results and checks

For a projectile that lands at its launch height, the vertical equation has a zero-time root and a positive flight-time root. The positive root gives the time in the air. The horizontal range is the horizontal displacement over that time. These same-height shortcuts do not apply unchanged when the landing elevation differs from the launch elevation; use the full vertical equation instead.

For an angled launch, set vertical velocity to zero to find the time to the peak. Substituting this time into the vertical position equation gives the rise above the launch point. The path is curved because horizontal position changes steadily while vertical position changes quadratically with time.

Check that position terms have units of metres and velocity terms have units of metres per second. Positive vertical velocity means upward motion in our convention; negative vertical velocity means downward motion. At

t=0

, the equations must return the stated initial position and velocity. With zero gravity, the model reduces to straight-line motion at constant velocity.

Use the full vertical position equation if launch and landing heights differ.
Range is horizontal displacement, not total path length.
Check the sign of vertical velocity and whether a time root describes the requested event.

4. Procedure for solving a projectile problem

Start by identifying the particle, observer, ground-fixed frame, origin, and positive directions. Sketch the stated projectile path and mark known launch conditions and the requested event. Resolve the initial velocity into components, keeping signs consistent with the axes.

List knowns and unknowns, then choose the vertical equation if the event is defined by height or vertical velocity. Solve for the physically relevant time. Substitute that time into the horizontal equation or velocity equations as needed. Keep units with substituted values, and round only after the calculation.

Finally, check that the result agrees with the motion: the projectile should move in the direction indicated by its velocity components, and a falling projectile should have downward acceleration even while it is rising. A same-height range should approach zero as launch speed approaches zero. These checks can reveal sign errors or a mistaken time root.

\Delta x=v_{0x}t

Solve the vertical condition for time when height or vertical velocity defines the event.
Use that same time to calculate horizontal position.
Verify initial conditions, units, signs, and physical direction.

Worked example

1. Horizontal launch from a platform

A small package leaves a platform horizontally at

8.00\ \mathrm{m/s}

from a height of

20.0\ \mathrm{m}

. Find its flight time and horizontal distance to the ground. Neglect air resistance.

Horizontal launch

The package starts with horizontal velocity and accelerates downward.

Set the frame and coordinates
The system is the package, observed from the stationary ground frame. Put the origin at the launch point, with $x$ positive horizontally toward the landing area and $y$ positive upward. The ground is at $y=-20.0\ \mathrm{m}$ . Initially, $v_{0x}=8.00\ \mathrm{m/s}$ and $v_{0y}=0$ .
$x_0=0,\quad y_0=0,\quad a_x=0,\quad a_y=-9.81\ \mathrm{m/s^2}$
Find the flight time
The landing condition is vertical position $y=-20.0\ \mathrm{m}$ . Use the vertical position equation; the positive root is the future landing time.
$-20.0=-\frac{1}{2}(9.81)t^2\quad\Rightarrow\quad t=2.02\ \mathrm{s}$
Find the horizontal distance
Horizontal speed is constant, so multiply it by the flight time. The resulting displacement is positive, matching the chosen direction.
$\Delta x=(8.00\ \mathrm{m/s})(2.02\ \mathrm{s})=16.2\ \mathrm{m}$

Answer: The package is in flight for

2.02\ \mathrm{s}

and lands

16.2\ \mathrm{m}

horizontally from the point directly below the platform edge.

Check: The vertical displacement is negative, as required for landing below launch height. The horizontal calculation reduces to metres. At release, the model gives the stated horizontal velocity and zero vertical velocity.

Worked example

2. Angled launch returning to launch height

A ball is launched at

20.0\ \mathrm{m/s}

at

30.0^\circ

above the horizontal and lands at its launch height. Find its flight time, horizontal range, and maximum rise above launch.

Angled launch

The velocity has horizontal and upward vertical components at launch.

Resolve the launch velocity
Use the ground-fixed frame, with the launch point as origin and upward positive. The ball’s initial velocity components follow from the given angle.
$v_{0x}=20.0\cos30.0^\circ=17.32\ \mathrm{m/s},\quad v_{0y}=20.0\sin30.0^\circ=10.0\ \mathrm{m/s}$
Find the time to the peak and total flight
At the peak, vertical velocity is zero. Since the landing height equals the launch height, the descent takes the same time as the ascent in this uniform-gravity model.
$t_{\mathrm{peak}}=\frac{10.0}{9.81}=1.02\ \mathrm{s},\quad t_{\mathrm{flight}}=2t_{\mathrm{peak}}=2.04\ \mathrm{s}$
Find range and maximum rise
Use the full flight time for horizontal displacement. For maximum rise, substitute the peak time into the vertical position equation.
$\Delta x=(17.32)(2.04)=35.3\ \mathrm{m},\quad y_{\mathrm{peak}}=(10.0)(1.02)-\frac{1}{2}(9.81)(1.02)^2=5.10\ \mathrm{m}$

Answer: The flight time is

2.04\ \mathrm{s}

, the horizontal range is

35.3\ \mathrm{m}

, and the maximum rise above launch is

5.10\ \mathrm{m}

.

Check: At the peak,

v_y=10.0-(9.81)(1.02)\approx0\ \mathrm{m/s}

. The horizontal component remains positive and constant. The rise is positive and has units of metres.

Worked example

3. Angled launch from an elevated ledge

A stone is launched from a ledge

12.0\ \mathrm{m}

above level ground at

14.0\ \mathrm{m/s}

and

40.0^\circ

above horizontal. Find when it reaches the ground and its horizontal distance from the launch point.

Launch from a ledge

The ground lies below the chosen launch-point origin.

Define coordinates and components
The system is the stone, observed from a stationary ground frame. Set the origin at the launch point, with positive $x$ to the right and positive $y$ upward; the ground is therefore at $y=-12.0\ \mathrm{m}$ . Resolve the initial velocity into components.
$v_{0x}=14.0\cos40.0^\circ=10.72\ \mathrm{m/s},\quad v_{0y}=14.0\sin40.0^\circ=9.00\ \mathrm{m/s}$
Solve the vertical landing condition
Use the vertical position equation with $y=-12.0\ \mathrm{m}$ . The negative root would describe a time before launch, so only the positive root is relevant.
$-12.0=9.00t-4.905t^2\quad\Rightarrow\quad t=2.73\ \mathrm{s}$
Calculate horizontal travel
The horizontal acceleration is zero, so use the same landing time with the constant horizontal velocity.
$\Delta x=(10.72\ \mathrm{m/s})(2.73\ \mathrm{s})=29.3\ \mathrm{m}$

Answer: The stone reaches the ground about

2.73\ \mathrm{s}

after launch, at a horizontal distance of

29.3\ \mathrm{m}

from the launch point.

Check: At landing,

y

is negative relative to the launch origin, consistent with the ground being lower. The stone is descending:

v_y=9.00-(9.81)(2.73)\approx-17.8\ \mathrm{m/s}

. The horizontal displacement is positive.

Common mistakes and how to avoid them

Using gravity as a horizontal acceleration or changing the horizontal velocity during flight.

Correction: In the stated model, gravity acts vertically, so

a_x=0

and horizontal velocity is constant.

Treating the launch speed as the horizontal velocity.

Correction: Resolve the launch velocity into components using the launch angle before applying the motion equations.

Using a same-height flight-time shortcut when the projectile lands at a different elevation.

Correction: Use the vertical position equation with the actual initial and final heights, then select the positive time.

Calling horizontal range the total distance travelled.

Correction: Range is horizontal displacement. The path length follows the curved trajectory and is a different quantity.

Lesson summary

Model the projectile as a particle in a ground-fixed frame, with air resistance neglected and gravity acting downward.
Use constant horizontal velocity and constant downward vertical acceleration.
Resolve the initial velocity into components and use the shared time to connect the two directions.
Check signs, units, initial conditions, the physical time root, and the predicted direction of motion.

Check your understanding

Question 1

A projectile is moving upward after launch. What is the direction of its acceleration in the uniform-gravity model?

Upward, because the projectile is rising
Downward, regardless of whether it is rising or falling
Horizontal, because its path is curved
Zero at the highest point

Show answer and explanation

Downward, regardless of whether it is rising or falling

Gravity provides constant downward acceleration throughout flight, including at the highest point.

Question 2

A projectile is launched horizontally at

6.0\ \mathrm{m/s}

. Neglecting air resistance, what is its horizontal velocity after

2.0\ \mathrm{s}

?

$0\ \mathrm{m/s}$
$6.0\ \mathrm{m/s}$
$19.6\ \mathrm{m/s}$
$-6.0\ \mathrm{m/s}$

Show answer and explanation

6.0\ \mathrm{m/s}

Horizontal acceleration is zero in this model, so horizontal velocity remains equal to its initial value.

Question 3

A projectile’s vertical velocity is

+4.0\ \mathrm{m/s}

at one instant. With upward positive, what is its vertical velocity

1.0\ \mathrm{s}

later?

$+13.8\ \mathrm{m/s}$
$+4.0\ \mathrm{m/s}$
$-5.8\ \mathrm{m/s}$
$-9.8\ \mathrm{m/s}$

Show answer and explanation

-5.8\ \mathrm{m/s}

Vertical velocity changes by

-gt

, giving

4.0-(9.81)(1.0)=-5.81\ \mathrm{m/s}

, which indicates downward motion.

Key terms

Projectile: A particle moving under the influence of gravity alone in the idealized model used here.
Uniform gravity: The model assumption that gravitational acceleration is constant in magnitude and direction during the motion.
Range: The horizontal displacement from launch to the specified landing point.
Component: The part of a vector along a chosen coordinate direction, such as $v_x$ or $v_y$ .

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2.2 · Model projectile motion under uniform gravity

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Separating horizontal and vertical motion to predict a particle’s path

What you will learn

1. Define the system and model

2. Governing equations and how to use them

3. Useful results and checks

4. Procedure for solving a projectile problem

1. Horizontal launch from a platform

2. Angled launch returning to launch height

3. Angled launch from an elevated ledge

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through EN PH 131

About this lesson and its review