Projectile Motion

From the groundFrom a cliff

Kicked off a cliff

• Review:• Objects Kicked off a cliff have:

X-dir Y-dirv= const. vi=0 m/sa= 0 m/s2 a = -9.8 m/sΔx= positive Δy= negative

vf= negative

Kicked off a cliff

• This is a combination of basicHorizontal Motion

&Vertical Motion

Projectile Motion- Level Ground

• Examples: Footballs, golf balls, bullets, catapults. There is no displacement in the Δy direction between the beginning “sea level” and the ending “sea level”

Projectile Motion- Level Ground

• This is a combination of Vertical Motion

&Horizontal Motion

Projectile Motion- Level Ground

• What do we know about:X-direction Y-direction

Projectile Motion- Level Ground

• What do we know about:X-direction Y-directionΔxright= + Δy= 0

v=constant vi= +

a= 0m/s2 vf = -

a= -9.8m/s2

Projectile Motion- Level Ground

• Attacking the problem:Initial velocities will be given in vectors:

25 m/s @ an angle of 30˚.

30˚

Projectile Motion- Level Ground

• Resolve the vector into components to determine Initial Velocity in the X and Y directions.

30˚

25 m/s

Projectile Motion- Level Ground

• Resolve the vector into components to determine Initial Velocity in the X and Y directions.

X-dir Y-dir

30˚

25 m/s

)30cos(2525

)30cos(

x

x

v

v

vx

vy )30sin(2525

)30sin(

y

y

v

v

Projectile Motion- Level Ground

• What do we know about:X-direction Y-directionΔxright= +vx Δy= 0

v=constant vi= +vy

a= 0m/s2 vf = -vy

a= -9.8m/s2Now you are ready to solve the problem…