PROJECTILE MOTION

RULESThere are only really a few rules:

All projectiles are freefalling objects

The horizontal motion is totally unaffected by the freefall motion!

Since there are no forces acting on it, horizontal motion is at constant speed!

AND NOW…

HORIZONTAL PROJECTILES•Horizontal Projectiles are easiest to work with•only formula used in horizontal (x) direction is:vx = dx / t

HORIZONTAL PROJECTILES•Horizontal Projectiles are the most basic•only formula used in horizontal (x) direction is:

vx = dx / t

constant speed!

HORIZONTAL PROJECTILES•vertical (y) direction is just freefall•all of the initial velocity is in the x direction•So,

€

v i• viy = 0

•since viy is in freefall,

• a = -9.8 m/s2€

t1

€

t2

€

t3

€

t4

start by drawing a picture:

EXAMPLEA person decides to fire a rifle horizontally at a bull’s-eye. The speed of the bullet as it leaves the barrel of the gun is 890 m/s. He’s new to the ideas of projectile motion so doesn’t aim high and the bullet strikes the target 1.7 cm below the center of the bull’s-eye.

What is the horizontal distance between the rifle and the bull’ s-eye?

label the explicit givens €

890 m /s

€

1.7 cm

givens (separate by direction):

EXAMPLEWhat is the horizontal distance between the rifle and the bull’ s-eye?

want:€

890 m /s

€

1.7 cm

X Y

€

vx =

€

dy =

€

ay =

€

v iy =€

890 m /s

€

−1.7 cm

€

−9.8 ms2

€

0 m /s

€

dx horizontal distance

€

1m100cm ⎛ ⎝ ⎜

⎞ ⎠ ⎟= −.017m

which equation do we use?

EXAMPLE

€

dy = v iyt +12

ayt2use to find time

rewrite equation for t

€

t =2dy

ay

€

/ 0

€

=2(−0.017m)

−9.8 ms2

€

= 0.059 s

Use t and vx to solve for dx

EXAMPLE

€

dx = vxt

€

=(890 ms )(0.059 s)

€

= 52.4 m

NON-HORIZONTAL PROJECTILES• vx is still constant• vy is still in freefall•only difference with non-horizontal is…•

€

v i€

v iy ≠ 0

NON-HORIZONTAL PROJECTILES•Angled Projectiles require a little work to get useful vi

•vi has an x and y component•need to calculate initial vx and vy

€

v i

BREAKING UP A VECTOR•every vector has 2 components to it•a horizontal component•a vertical component

•they add up to the total€

vx

€

vy

€

v

BREAKING UP A VECTOR

SOHCAHTOA

€

θ

€

=v cosθ

€

=v sinθhypotenuse

adjacent

opposite€

sin = oppositehypotenuse

€

cos = adjacenthypotenuse

€

sinθ = OH

€

=vy

v

€

⇒ vy = v sinθ

€

vx

€

vy

€

v

BREAKING UP A VECTOR

SOHCAHTOAneed to find θ?

€

tanθ = OA

€

θ =tan−1 OA ⎛ ⎝ ⎜

⎞ ⎠ ⎟

hypotenuse

adjacent

opposite

NON-HORIZONTAL PROJECTILES•need to calculate initial vx and vy

€

v

€

vx€

vy

€

θ

€

=v cosθ€

=v sinθ

RELIEF

VISUALIZING PROJECTILES•first enter vectors•focus on vx

vx is constant the whole flight!

VISUALIZING PROJECTILES•first enter vectors•focus on vx

•focus on vy

vy decreases as it rises!

by how much per second?

no vy at the top!

VISUALIZING PROJECTILES

VARIED ANGLES•which projectile angle shoots highest?• larger θ means faster viy•which projectile angle shoots farthest?•45° has perfect balance of fast vx and long flight time.

LET’S ANALYZE THE JUMP

Try Graphing ItGo to pg. 331