PHYS 20 LESSONS

Unit 2: 2-D Kinematics

Projectiles

Lesson 5: 2-D Projectiles

Reading Segment #1:

Horizontal Projectiles

To prepare for this section, please read:

Unit 2: p.18

E. 2-D Projectiles

When a projectile is thrown, the path of the object is

a parabola.

i.e.

Animation:

Pirate Ship

http://library.thinkquest.org/27585/lab/sim_pirates.html

To study 2-D projectile motion, we will analyze two situations:

1. Projectiles launched horizontally

2. Projectiles launched at an angle

We will assume there is no air resistance for all examples.

E1. Projectiles launched horizontally

Consider a cannonball fired horizontally off of a cliff.

We will analyze this motion from the horizontal (x)

and the vertical (y) perspectives separately.

Horizontal (x) perspective:

Because there is no air resistance, there is no horizontal

acceleration. It will move with a constant horizontal velocity

(much like a cannonball that experiences no gravity at all).

Vertical (y) perspective:

It has no vertical velocity at the start,

but it is accelerating downward at

9.81 m/s2. So, it will speed up vertically

(from rest).

Animation:

Horizontally-Launched Projectile

http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/vectors/hlp.html

Summary:

Horizontal Motion

Moves at a constant velocity

Use the formula v = d

t

Vertical Motion

Starts from rest (vi = 0).

Accelerates at 9.81 m/s2 downward.

Use the kinematic equations.

Ex. 1 An object is launched horizontally off of a cliff that

is 18.0 m high. How far does it travel horizontally

before it hits the ground.

Assume no air resistance and level ground.

14.0 m/s

18.0 m

d ?

We will analyze the vertical motion first.

From this perspective, it falls (from rest) a height of 18.0 m.

Vertical (y): Ref: Down +

Up -

vi = 0

vf

a = 9.81 m/s2

d = 18.0 m

t ?

We need to find the time of the flight.

Use the equation d = vi t + 0.5 a t2

d = vi t + 0.5 a t2

d = t2

0.5 a

t = d = 18.0 m

0.5 a 0.5 (9.81 m/s2)

= 1.916 s

We can now use this flight time for the horizontal motion.

Horizontal (x) Motion:

It moves at a constant horiztonal speed of 14.0 m/s.

So, v = d

t

d = v t

= (14.0 m/s) (1.916 s)

= 26.8 m

Practice Problems

Try these problems in the Physics 20 Workbook:

Unit 2 p. 19 #1 - 7

Reading Segment #2:

Diagonal Projectiles

To prepare for this section, please read:

Unit 2: p.7

E2. Projectiles launched diagonally

Consider a cannonball fired diagonally.

v

We will analyze this motion from the horizontal (x)

and the vertical (y) perspectives separately.

Horizontal (x) perspective:

v

vx

The horizontal component of the velocity (vx)

will remain constant throughout the motion.

There is no horizontal acceleration.

Vertical (y) perspective:

v

Vertically, the object will accelerate downward at 9.81 m/s2.

Thus, its speed will decrease on the way up,

and its speed will increase on the way down.

Vertical (y) perspective:

v

Notice, at the top of the motion (max height),

although the vertical component is zero, it is not at rest.

There is still a horizontal component.

Animation:

Diagonally-launched projectile:

http://www.glenbrook.k12.il.us/gbssci/phys/mmedia/vectors/nhlp.html

Summary (Diagonal Projectiles)

Horizontally:

Moves at a constant velocity. Use vx = d / t

Vertically:

Accelerates downward at 9.81 m/s2

Recall (symmetry)

At the same height, vup = vdown

If it returns to the same height, tup = tdown

Ex. 2 A golf ball is hit with a speed of 78.0 m/s at 13.0 above

the horizontal. Find:

a) the maximum horizontal distance the ball travels

while it is in the air

b) the maximum height of the ball

Assume no air resistance and level ground.

Step 1: Find the x- and y-components of the initial velocity

y 78.0 m/s

vy

13.0

vx x

7813cos xv

13cos78xv

= 76.00 m/s

7813sin yv

13sin78yv

= 17.546 m/s

a) Vertical (y) perspective:

From the vertical perspective, the ball starts with an

upward velocity of 17.546 m/s.

17.546 m/s

It goes up and then comes back

down to the same height.

17.546 m/s

It will have the same speed when

it returns to the same height.

a) Vertical (y) perspective: Ref: Up +

List Down -

vi = +17.546 m/s

vf = -17.546 m/s

a = -9.81 m/s2

d =

t = ? 17.546 m/s

We need to find the total time

the ball is in the air.

17.546 m/s

a) Vertical (y) perspective: Ref: Up +

List Down -

vi = +17.546 m/s

vf = -17.546 m/s

a = -9.81 m/s2

d =

t = ?

Equation: a = vf - vi

t

a = vf - vi

t

a t = vf - vi

t = vf - vi

a

= (-17.546 m/s) - (+17.546 m/s)

-9.81 m/s2

= 3.5772 s

Horizontal (x) Perspective:

The ball travels at a constant speed of 76.00 m/s (vx)

for a total time of 3.5772 s.

vx = d

t

d = vx t

= (76.00 m/s) (3.5772 s)

= 272 m

b) Vertical (y) perspective:

When the ball reaches its maximum height, its vertical

velocity is zero (even though it is still moving horizontally).

Rest

17.546 m/s

b) Vertical (y) perspective: Ref: Up +

List Down -

vi = +17.546 m/s

vf = 0 Rest

a = -9.81 m/s2

d =

t = ? 17.546 m/s

Equation: vf2 = vi

2 + 2 a d

vf2 = vi

2 + 2 a d

vf2 - vi

2 = 2 a d

d = vf2 - vi

2

2a

= 0 - (17.546 m/s)2

2 (-9.81 m/s2)

= 15.7 m

Additional Animations:

Cannon (target):

http://zebu.uoregon.edu/nsf/cannon.html

Diagonal Projectiles:

http://www.hazelwood.k12.mo.us/~grichert/projectile/proj.html

http://www.msu.edu/user/brechtjo/physics/cannon/cannon.html

Practice Problems

Try these problems in the Physics 20 Workbook:

Unit 2 p. 22 #1 - 5