Projectile Motion

What is a projectile motion?

A projectile motion is a two-dimensional

motion with the force of gravity acting upon

it.

TERMS

Projectile

It is an object thrown by the exertion of a force, having its

own trajectory.

Trajectory

It is the curved path taken by a projectile.

Time of Flight

It is the duration in which the projectile is in flight.

Range It is the horizontal

displacement from the takeoff point to the landing

point.

Ballistics

It is the study of projectile motion.

Since projectile motion is a two-dimensional motion, it

has a horizontal and vertical motion. The horizontal

motion is constant, while the vertical motion is

constantly changing due to the force of gravity.

Projections

Horizontal Projection

In the horizontal projection, the projectile

doesn’t encounter an upward motion. It is directed horizontally

straight and goes downward until it reaches

a surface.

The Three Kinematic Equations

D = displacementa = acceleration

t = timevf = final velocityvi = initial velocity

Solving the Horizontal Projection

Determine the horizontal and vertical components

of the problem.

HorizontalX = horizontal displacement

Ax = horizontal acceleration (0 m/s2)

T = timeVfx = final horizontal velocity

Vix = initial horizontal velocity

Vertical

y = vertical displacementAy = vertical acceleration (-9.8

m/s2) T = time

Vfy = final vertical velocity

Viy = initial vertical velocity (0 m/s2)

Draw the trajectory.

Identify the unknown.

Solve the missing by manipulating the three kinematic equations.

Horizontal

Vertical

Other Equations

Trajectory

𝑦=12𝑔 [ 𝑥𝑣 𝑖 ]

2

Velocity at any instant of time

Time of Flight

𝑡=√ 2 𝑦𝑔

Range

𝑟=𝑣 𝑖𝑥 √ 2 𝑦𝑔

Sample Problems

A soccer ball is kicked horizontally off a 22-

meter high hill and lands a distance of 35 m from

the edge of the hill. Determine the initial

horizontal velocity of the soccer ball.

Fred throws a baseball 42 m/s horizontally from a

height of 2 m. How far will the ball travel before it reaches the ground?

Solutions

Given

Horizontal Vertical

x = 35 m y = -22 m

vix = ? viy = 0 m/s

Vfx = ? Vfy = ?

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

t = ? t = ?

Find t.

Solve for vix

Given

Horizontal Vertical

x = ? y = -2 m

vix = -42 m/s viy = 0 m/s

Vfx = -42 m/s Vfy = ?

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

t = ? t = ?

Find t.

Solve for x

Angular Projection

In the angular projection, the projectile encounters

an upward motion, therefore the maximum height is needed to be

solved.

The same kinematic equations are to be used in

solving the problem, but what differs is the initial

velocity of both horizontal and vertical, and also the maximum height of the

trajectory.

Solving the Angular

Projection

Horizontal Initial Velocity

𝑣 𝑖𝑥=𝑣 𝑖cos𝜃

Vertical Initial Velocity

𝑣 𝑖𝑦=𝑣 𝑖 sin𝜃

Other Equations

Velocity at any instant of time

Time of Flight

𝑇=2 (𝑣 𝑖𝑦 )𝑔

Range

𝑟=𝑣 𝑖𝑥 √ 2 𝑦𝑔

Maximum Height

𝐻=𝑣 𝑖𝑦𝑇2−12𝑔(𝑇2 )

2

Sample Problems

A long jumper leaves the ground with an initial

velocity of at an angle of above the ground.

Determine the time of flight, the horizontal distance, and the peak height of the long-

jumper.

A cannonball was shot 60 m from a cliff at an angle of . It has an initial velocity of . Find the maximum height,

time of flight and the horizontal displacement of

the cannonball.

Solutions

Given

Horizontal Vertical

x = ? y =

vix = ? viy = ?

Vfx = ? Vfy = ?

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

t = ? t = ?

Find vix

Find viy

Horizontal Vertical

x = ? y =

vix = 10.6 m/s viy = 5.63 m/s

Vfx = 10.6 m/s Vfy = -5.63 m/s

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

t = ? t = ?

Find t.

Solve for x

Solve for y

Given

Horizontal Vertical

x = ? y = ?

vix = ? viy = ?

Vfx = ? Vfy = ?

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

t = ? t = ?

Find vix

Find viy

Horizontal Vertical

x = ? y = ?

vix = 15.05 m/s viy = 19.97 m/s

Vfx = 15.05 m/s Vfy = ?

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

t = ? t = ?

Find t from the launch to the maximum height

1

0 1

1

Solve for y

Find the freefall t2

16. 4

Solve for x