Quiz

• 1. An object is dropped from a height of 6.5 meters. How long does it take to reach the ground?

• 2. An object is moving at a constant velocity of 5.7 m/s. How far does the object travel in 1.3 seconds?

Projectile Motion

Independent Vector Analysis

Objectives

1. Recognize that the vertical and horizontal motions of a projectile are independent.

2. Relate the height, time in the air and initial vertical

Terms

1. Projectile Any object that moves through the air

2. Trajectory The path a projectile takes through space

3. Range The horizontal distance a projectile moves from the launching point

What is Projectile Motion?

is given an initial thrust,When a projectile

and we ignore air resistance,it moves through the air

with only what acting on it?

Gravity!

Bullet Fired vs. Bullet Droped

• Mythbusters Youtube• http://www.youtube.com/watch?v=abUBrQmI

33Q• Draw it in your notes• Conclusion – Vertical Motion (in the y-

direction) is accelerated because of gravity.

Packaged Dropped from Plane• https://d3jc3ahdjad7x7.cloudfront.net/

XLzJ3XwhhuLdhQ0hZKkuoSV8JFceSEDG5LAsmIowJxNz2tsu.jar

• Draw it in your notes. • Click on the red x to drop the package.• During its decent, press pause at several

different times.• Where is the package, relative to the plane, as

it falls to the ground?• Conclusion – Horizontal Motion (in the x-

direction) is constant velocity because• Horizontal acceleration = ax = 0 = zero

How Does Projectile Motion Work?The vertical an horizontal motion of a projectile

are independent of each other

How Does Projectile Motion Work?

We know that the horizontal motion of a launched ball does not affect its vertical motion.

A projectile launched horizontally has no initial vertical velocity

The vertical and horizontal motion of a projectile are independent of each other

Vi = 0 m/s

How Does Projectile Motion Work?

Therefore a horizontally launched projectile has a vertical motion just like an object in freefall.

And horizontal motion just like an object in constant motion.

How Does Projectile Motion Work?

Therefore a horizontally launched projectile has a vertical motion just like an object in freefall.

And horizontal motion just like an object in constant motion.

Problem Solving Strategies1. Motion in two dimensions can be solved by breaking the

problem into two interconnected one-dimensional problems.

x

yR

vi

a

t

vf

vf = vi + at

vi a t vf

vf = vi + at

Problem Solving Strategies

• Projectile motion can be divided into a vertical motion problem and a horizontal motion problem.

The vertical component

can be solved like a freefall

problem

d = vit + ½ at2

vi

a

t

vf

d

The horizontal component can be solved like a

constant motion problem d

v = d/t

v t

Problem Solving StrategiesBoth the vertical and horizontal components of a projectile are connected by time (t).

The time for a projectile to hit the ground is the same for the vertical component and the horizontal component.

tx

tytP

tx = ty = tR

Example 1A stone is thrown horizontally at a speed of 5.0 m/s

from the top of a cliff 78.4 m high. How long does it take the stone to reach the bottom of the cliff?

What is Step 1?Draw a picturevx = 5 m/s

dy = 78.4 m

vi = 0 m/s

g = -9.8 m/s2

vf = ?

ty = ?

dx = ? tx = ?

ty = txtP=

Example 1 (continued)A stone is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 78.4 m

high. How long does it take the stone to reach the bottom of the cliff?

G:

U:

E:

S:

S:

Horizontal VerticalVx = 5.0 m/s

dx = ?tx = ?

dy = 78.4 mvi = 0 m/sa = -9.8 m/s2

ty = ?vf = ?

d = vit + ½ at2

d = ½ at2

t = 2d a

t =2 (78.4 m)(-9.8m/s2)

t = 4.0 s

Example 2A stone is thrown horizontally at a speed of 5.0 m/s from

the top of a cliff 78.4 m high. How far from the cliff does the stone hit the ground ?

G:

U:

E:

S:

S:

Horizontal Vertical

vx = 5.0 m/s

dx = ?tx = ?

dy = 78.4 mvi = 0 m/sa = -9.8 m/s2

ty = ?vf = ?

d = vit + ½ at2

d = ½ at2

t = 2d a

t =2 (78.4 m)(-9.8m/s2)

t = 4.0 s

t = 4.0 s

v = d/t

d = vt

d = (5.0 m/s)●(4.0 s)

d = 20 m

Example 3A steel ball rolls with a constant velocity across a tabletop 0.950 m high. It

rolls off the table and hits the ground 0.352 m from the edge of the table. How fast was the ball rolling just as it left the table?

vx = ?

dy = 0.950 m

vi = 0 m/s

g = -9.8 m/s2

vf = ?

ty = ?

dx = 0.352 mtx = ?

Example 3 (continued)

G:

U:

E:

S:

S:

Horizontal Verticaldx = 0.352 m

tx = ?vx = ?

dy = 0.950 mvi = 0 m/sa = -9.8 m/s2

ty = ?vf = ?

d = vit + ½ at2

d = ½ at2

t = 2d a

t =2 (0.950m)(-9.8m/s2)

t = 0.44 s

v = d/t

vx = (0.352 m) / (0.44 s)

vx = 0.8 m/s

A steel ball rolls with a constant velocity across a tabletop 0.950 m high. It rolls off the table and hits the ground 0.352 m from the edge of the table. How fast was the ball rolling just as it left the table?

= 0.44 s

Example 4Divers at Acapulco dive horizontally from a cliff that is 65 meters high. If the rocks below the cliff protrude 27 meters beyond the edge of the cliff, what is the minimum horizontal velocity needed to safely clear the rocks below?

vx = ?

dy = 65 m

vi = 0 m/s

g = -9.8 m/s2

vf = ?

ty = ?

dx = 27 mtx = ?

Example 4 (continued)

G:

U:

E:

S:

S:

Horizontal Verticaldx = 27 m

tx = ?vx = ?

dy = 65 mvi = 0 m/sa = -9.8 m/s2

ty = ?vf = ?

d = vit + ½ at2

d = ½ at2

t = 2d a

t =2 (65 m)(-9.8m/s2)

t = 3.64 s

v = d/t

vx = (27 m) / (3.64 s)

vx = 7.43 m/s

= 3.64 s

Divers at Acapulco dive horizontally from a cliff that is 65 meters high. If the rocks below the cliff protrude 27 meters beyond the edge of the cliff, what is the minimum horizontal velocity needed to safely clear the rocks below?

Summary• Projectile Motion is calculated using independent

equations.

• The horizontal movement of a projectile resembles constant motion.

• The vertical movement of a projectile resembles freefall.

• Time is the same for both vertical and horizontal components of projectile motion.