Horizontally launched projectiles

Math of Horizontal Projectiles

Unit 4

2-D Motion

Lesson 5

01.

02.

03.

PHYSICS | LESSON 4.5 | MATH OF HORIZONTAL PROJECTILES

How do the horizontal and vertical velocities, accelerations and displacements change for a projected object?

How does a projected object’s motion change from its
initial launch, to its maximum height and its final position?

What are the initial conditions for a horizontally or angle launched projectile?

Essential Questions

2

LEARNING
OBJECTIVES

Analyze and distinguish between horizontal and
vertical components of a projectile in two dimensions.

Identify the initial conditions of a horizontally launched
projectile.

01.

02.

Main
Ideas

Theinitial vertical velocity
of a horizontally launched
projectile is 0 m/s.

Thehorizontal motion of a
projectile is INDEPENDENT
of the vertical motion.

As you can see, we can prove mathematically
that the horizontal velocity stays constant! Please copy for your notes.

Horizontal

Vertical

ax = 0 m/s^2

ay = -9.8 m/s^2

Vf,x= v0,x

vf,y = v0,y

+ ayt

∆x = v0,xt

∆y = ½ayt^2 + v0,yt

Kinematics in 2D

The horizontal displacement or RANGE
EQUATION is a simple equation with only 3
variables: range, initial x-velocity and time.

Horizontal

Vertical

ax = 0 m/s^2 ay = -9.8 m/s^2

vf,x= v0,x

vf,y = v0,y

+ ayt

∆x = v0,xt

∆y = ½ayt^2 + v0,yt

Kinematics in 2D

​Range equation

This is another equation you’ve used before! Don’t forget
that you can only use vertical components to solve for an
unknown variable in the height equation.

Horizontal

Vertical

ax = 0 m/s^2 ay = -9.8 m/s^2

vf,x = v0,xvf,y = v0,y

+ ayt

∆x = v0,xt

∆y = ½ayt^2 + v0,yt

🡨Height Equation

Kinematics in 2D

Time is the only variable that can be used in both dimensions.

Horizontal

Vertical

ax = 0 m/s2 ay = -9.8 m/s2

vf,x = v0,x

vf,y = v0,y

+ ayt

∆x = v0,xt

∆y = ½ayt2 + v0,yt

Kinematics in 2D

Multiple Choice

Why can we use time in multiple dimensions?

• Time can only be measured in a single dimension.
• Time is a vector.

• Time is irrelevant.

• Time is a scalar

Vertically, the object undergoes constant acceleration due to gravity.
9.81 m/s^2

Kinematics in 2D

Horizontally, the object experiences no
acceleration (constant velocity).

Kinematics in 2D

One VERY important thing to remember that only
pertains to horizontally projected objects is that the
initial y-velocity is zero!!!

Kinematics in 2D

A baseball is thrown
horizontally at 33.5 m/s.

At 0.55 sec, how far
horizontally has it
moved?

Baseball Problem

Match

Match the variables

• Range equation

• 33.5 m/s

• 0.55 s

• Unknown

• 9.81 m/s²

v_x

t

x

g

• $$\Delta x=v_{x,0}t$$

• v_x

• t

• x

• g

Use the Range equation to find the answer.

Baseball Problem

PHYSICS | LESSON 4.5 | MATH OF HORIZONTAL PROJECTILES

29

Use the Range equation to find the answer.

9. ∆x = v0,xt 🡪 ∆x = 33.5(0.55) =

Baseball Problem

Multiple Choice

A baseball is thrown horizontally at 33.5 m/s. At 0.55 sec, how far horizontally has it moved?

• 15.2 m
• 18.4 m

• 22.1 m
• 25.0 m

Use the Range equation to find the answer.

9. ∆x = v0,xt 🡪 ∆x = 33.5(0.55) = 18.4 m

Baseball Problem

How far vertically has it fallen?

Use the Height equation but remember; the initial y-velocity of a horizontally projected object is zero!!

Baseball Problem

34

Use the Height equation, the initial y-velocity of a
horizontally projected object is zero!!

∆y = ½ayt2 + v0,yt

Baseball Problem

​0

Multiple Choice

A baseball is thrown horizontally at 33.5 m/s. At 0.55 sec, how far vertically has it fallen?

• 2.10 m
• 0.25 m
• 1.48 m
• 1.00 m

∆y = ½ayt2 + v0,yt

yf – y0 = ½ayt2 + v0,yt

0 – y0 = ½(-9.8)0.552 + 0(t)

y0 = ½(-9.8)0.552

y0 = 1.48 m

Baseball Problem

P

What is its horizontal velocity at 0.55 s?

Baseball Problem

Multiple Choice

A baseball is thrown horizontally at 33.5 m/s. At 0.55 sec, what is its horizontal velocity?

• 30 m/s
• 33.5 km/h
• 33.5 m/s
• 0 m/s

What is its horizontal velocity
at 0.55 s?

TRICK QUESTION!

The horizontal velocity always stays constant, so it
is the same as the initial x-velocity, 33.5 m/s.

Baseball Problem

What is its vertical velocity at 0.55 s?

Baseball Problem

12. What is its vertical velocity at 0.55 s?

Use vy,f = v0,y

+ ayt

Baseball Problem

​0

Multiple Choice

A baseball is thrown horizontally at 33.5 m/s. At 0.55 sec, what is the vertical velocity?

• -3.20 m/s

• -7.80 m/s

• -10.5 m/s

• -5.39 m/s

12. What is its vertical velocity at 0.55 s?

Use vy,f = v0,y

+ ayt

vy,f = 0 + (-9.8)0.55
vy,f = -5.39 m/s

It’s negative because it’s moving downward.

Baseball Problem

PHYSICS | LESSON 4.5 | MATH OF HORIZONTAL PROJECTILES

50

Step 1 - Time is the only variable to cross over between
the horizontal and vertical sides of the chart.

Step 2 - Use whichever column has more information to
solve for time.

Horizontal Projectile
Problem Solving Strategy

If time is given, you need to analyze the given variables and figure out which equation you have to solve first.

You have a 1/3 chance of getting it right.

Horizontal Projectile Practice

PHYSICS | LESSON 4.5 | MATH OF HORIZONTAL PROJECTILES

57

• The initial y-velocity of a horizontal projectile is

zero.

• Time in the x and y direction is the same for a

projectile.

Let’s Summarize

• The initial y-velocity of a horizontal projectile is

zero.

• Time in the x and y direction is the same for a

projectile.

• The range of the projectile is the horizontal displacement.

• The negative sign for displacement and velocity is due to the downward direction.

Let’s Summarize