Motionin Two Dimensions

Chapter 7.2

Projectile Motion

What is the path of a projectile as it moves through the air? Parabolic? Straight up and down?

Yes, both are possible.

What forces act on projectiles? Only gravity, which acts only in the negative

y-direction. Air resistance is ignored in projectile motion.

Choosing Coordinates & Strategy

For projectile motion: Choose the y-axis for vertical motion where

gravity is a factor. Choose the x-axis for horizontal motion. Since

there are no forces acting in this direction (of course we will neglect friction due to air resistance), the speed will be constant (a = 0).

Analyze motion along the y-axis separate from the x-axis.

If you solve for time in one direction, you automatically solve for time in the other direction.

The Trajectory of a Projectile

•What does the free-body diagram look like for force?

Fg

The Vectors of Projectile Motion

What vectors exist in projectile motion? Velocity in both the x and y directions. Acceleration in the y direction only.

vy (Increasing)

vx (constant)

ay

ax = 0

Why is the velocity constant in the x-direction?

•No force acting on it.

Why does the velocity increase in the y-direction?

•Gravity.

Formulas for Motion of ObjectsFormulas for Motion of Objectsassuming d is displacement from origin and time starts at 0.assuming d is displacement from origin and time starts at 0.

Equations to use when Equations to use when an accelerating object an accelerating object has an initial velocity.has an initial velocity.

Form to use when Form to use when accelerating object accelerating object starts from rest (vstarts from rest (v ii = 0). = 0).

d = ½ (vd = ½ (vi i + v+ vff) t = v) t = vaveavett d = ½ vd = ½ vff t t

vvff = v = vii + at + at vvff = at = at

d = vd = vii t + ½ a(t) t + ½ a(t)22 d = ½ a(t)d = ½ a(t)22

vvff22 = v = vii

2 2 + 2ad+ 2ad vvff22 = 2ad = 2ad

There are only so many things a projectile can do

d = vave tvf = vi + atvave = ½(vf+vi)vf

2=vi 2+2ad

d = ½ at2

vf = vi + atvf

2=vi 2+2ad

Note: vi<>0 is possible

Solve x and y separately. t is when the object stops.

d = vi t + ½ at2

vf = vi + atvf

2=vi 2+2ad

Beware cases where t <> hang time

Solve x and y separately. t is when the object hits the ground.

Solve x and y separately. t is when the object hits the ground.

Move horizontally

Fall Go up and down

Be shot at an angle

Be shot at an angle

Be shot horizontally

Ex. 1: Launching a Projectile Horizontally

A cannonball is shot horizontally off a cliff with an initial velocity of 30 m/s. If the height of the cliff is 50 m: How far from the base of the cliff does

the cannonball hit the ground? With what speed does the cannonball hit

the ground?

Note: This is the same problem as “a life raft is dropped from an airplane …”

Diagram the problem

50mFg = Fnet a = -g

vi

vf = ?

vx

vyx = ?

State the Known & Unknown

Known: xi = 0 vix = 30 m/s yi = 0 viy = 0 m/s

a = -g y = -50 m

Unknown: x at y = -50 m vf = ?

Perform Calculations (y)

y-direction: vy = -gt y = viyt – ½ gt2

Using the first formula above: vy = (-9.8 m/s2)(3.2 s) = 31 m/s

g

yt

2

ssm

mt 2.3

/81.9

5022

Perform Calculations (x)

x-Direction x = vixt x = (30 m/s)(3.2 s) = 96 m from the

base. Using the Pythagorean Theorem:

v = vx2 + vy

2

v = (30 m/s)2 + (31 m/s)2 = 43 m/s

Ex. 2: Projectile Motion above the Horizontal

A ball is thrown from the top of the Science Wing with a velocity of 15 m/s at an angle of 50 degrees above the horizontal.

What are the x and y components of the initial velocity? What is the ball’s maximum height? If the height of the Science wing is 12 m, where will the

ball land?

Diagram the problem

y

x

vi = 15 m/s

= 50°

Fg = Fnet a = -g

Ground

12 m

x = ?

vi = 15 m/s

vix

viy

= 50°

State the Known & Unknown

Known: xi = yi = vi = = a =

Unknown: ymax = ? t = ? x = ? viy = ? vix = ?

Hor Vert

Vix = Viy =

Vfx = Vfy =

dx = dy =

ax = ay =

t = t =

=

State the Known & Unknown

Known: xi = 0 yi = 12 m vi = 15 m/s = 50° a = -g

Unknown: ymax = ? t = ? x = ? viy = ? vix = ?

Perform the Calculations (ymax)

y-direction: Initial velocity: viy = visin

viy = (15 m/s)(sin 50°) viy = 11.5 m/s

Time when vfy = 0 m/s: vfy = viy + at = viy – gt 0 = viy – gt viy = gt t = viy / g t = (11.5 m/s)/(9.81 m/s2) t = 1.17 s

Determine the maximum height: d = vit + ½ at2

dmax = viyt – ½ gt2, plus 12 m ymax = 12 m + (11.5 m/s)(1.17 s) – ½ (9.81 m/s2)(1.17 s)2

ymax = 18.7 m

vi = 15 m/s

vxi

vyi

= 50°

Perform the Calculations (t)

Since the ball will accelerate due to gravity over the distance it is falling back to the ground, the time for this segment can be determined as follows

Time when ball hits the ground: ymax = viyt – ½ gt2

Since yi can be set to zero as can viy, t = 2*ymax/g t = 2(18.7 m)/ (9.81 m/s2) t = 1.95 s

By adding the time it takes the ball to reach its maximum height to the time it it takes to reach the ground will give you the total time.

ttotal = 1.17 s + 1.95 s = 3.12 s

Perform the Calculations (x)

x-direction: Initial velocity: vix = vicos

vix = (15 m/s)(cos 50°)

vix = 9.64 m/s

Determine the total distance: x = vixt x = (9.64 m/s)(3.12 s) x = 30.1 m

vi = 15 m/s

vxi

vyi

= 50°

Analyzing Motion in the x and y directions independently.

x-direction: ddxx = v = vix ix t = vt = vfxfxtt vvixix = v = viicoscos

y-direction:y-direction: ddyy = ½ (v = ½ (vii + v + vff) t = v) t = vavgavg t t vvff = v = viyiy + gt + gt ddyy = v = viyiy t + ½ g(t) t + ½ g(t)22

vfy2 = viy

2 + 2gdd vviyiy = v = viisinsin

Key Ideas

Projectile Motion: Gravity is the only force acting on a

projectile. Choose a coordinate axis that where

the x-direction is along the horizontal and the y-direction is vertical.

Solve the x and y components separately.

If time is found for one dimension, it is also known for the other dimension.