Projectile Motion

• A projectile is any object in free fall near the surface of the Earth

• A projectile moves horizontally as well as vertically

• Examples:– A baseball, football,

basketball, or soccer ball in flight

– A bullet fired from a gun

– A marble that has rolled off the table

– A human long-jumper

• The path of a projectile is a parabola

• Components simplify projectile motion:1) Displacements, velocities, and accelerations

are resolved into x- and y-components

x

y

V

x

y

Vy

Vx

2) The 1D-motion equations are applied to x- and y-directions independently

No. Equation

1 x = ½(vx0 + vx)t

2 vx = vx0 + axt

3 x = vx0t + ½ axt2

4 vx2 = vx0

2 + 2axx

No. Equation

1 y = ½(vy0 + vy)t

2 vy = vy0 + ayt

3 y = vy0t + ½ ayt2

4 vy2 = vy0

2 + 2ayy

x-direction:

y-direction:

3) The x- and y-components are recombined to determine the resultant motion

x

y

Vy

Vx x

y

V

X-Direction in Projectile Motion

• We assume air resistance is zero• Horizontal motion is constant velocity

vx = vx0 = constant

x = vxt

• Constant velocity means acceleration is zero

ax = 0

X1

X2

X3

Y-Direction in Projectile Motion

• We assume air resistance is zero

• Vertical motion is constant acceleration, ay = -g = -9.80 m/s2

y = ½(vy0 + vy)t

vy = vy0 - gt

y = vy0t - ½ gt2

vy2 = vy0

2 - 2gy

Y1

Y2

Y3

Y4

Projectile Launched Horizontally• Given:

vx = constant = vx0

vy0 = 0y = -h

• Unknowns:t = ?x = ?v = ? vx

h

v

• How long will it be airborne? Use Eqn. Y3:

-h = -½ gt2

t = (2h/g)½

• How far will it go? Use Eqn. X2:

x = vxt

• How fast will it hit? Use Eqn. Y4:

vy2 = 2gh

v2 = vx2 + vy

2

v2 = vx2 + 2gh

H1

H2

H3

• Example:Assume Wile E. Cayote fires the cannonball

horizontally at 25.2 m/s at a height of 7.91 m:

t = (2h/g)½ = [2(7.91 m)/(9.80 m/s2)]½ = 1.27 s

x = vxt = (25.2 m/s)(1.27 s) = 32.0 m

v = (vx2 + 2gh)½

= [(25.2m/s)2 + 2(9.80m/s2)(7.91m)]½ = 28.1 m/s

Projectile Launched at an Angle• Given:

vx = v0cos

vy0 = v0sin y = 0

• Unknowns:t = ?x = ?v = ?h = ?

v0

h

vx

v0vy0

vx

• How long will it be airborne? Use Eqn. Y3:

0 = v0sin·t – (½)gt2

t = (2v0/g)sin• How far will it go? Use Eqn. X2:

x = vxt = (v0cos)t

x = (2v02/g)cossin

• How fast will it hit? Use Eqn. Y1:

0 = ½(vy0 + vy)t

vy = -vy0

v = v0 (in magnitude)

A1

A3

A2

• How high will it go? Use Eqn. Y4 (half way):

So y = h and vy = 0

vy2 = vy0

2 - 2gy

0 = vy02 - 2gh

h = vy02/(2g)

h = (v0sin)2/(2g)

V0

h

A4

• Example:A golf ball is hit at an initial velocity of 43.9 m/s at

an angle of 33.3 from horizontal:

vx = v0cos = (43.9 m/s)cos(33.4) = 36.6 m/s

vy0 = v0sin = (43.9 m/s)sin(33.4) = 24.2 m/s

t = (2v0/g)sin = [2(43.9 m/s)/(9.80 m/s2)]sin(33.4) = 4.93 s

x = vxt = (2v02/g)cossin

= [2(43.9 m/s)2/(9.80 m/s2)]cos(33.4)sin(33.4)= 181 m

h = vy02/(2g) = (v0sin)2/(2g)

= [(43.9 m/s)sin(33.4)]2/(9.80 m/s2)/2 = 29.8 m