PROJECTILE MOTION

IMPORTANT KINEMATICS EQUATIONS!

vf = vi + atd = vit + ½ at2

vf2 = vi

2 + 2ad

¡ vf = final velocity¡ vi = initial velocity¡ a = acceleration¡ t = time¡ d = displacement

WHAT IS PROJECTILE MOTION?¡ Projectile motion is an example of 2-dimensional motion

¡ Projectile: object upon which the only force acting on it is gravity¡ Ex: object that is fired, thrown, or shot near the Earth’s surface,

and is in free-fall¡ Trajectory: a projectile’s path through space

¡ Trajectory of a projectile in free fall = parabola

We will ignore the effects of air

resistance on the projectile

¡ An object projected horizontally will reach the ground in the same time as an object dropped vertically§ No matter how large the

horizontal velocity is, the downward pull of gravity is always the same

0 svox

Vertical motion is the same for each ball

3 s

2 s

1 s

Consider horizontal and vertical motion separately:

Compare Displacements and Velocities

0 s 0 s1 svox 2 s 3 s

1 svy2 s

vx

vy3 s

vx

vy

Horizontal velocity doesn’t change.

Vertical velocity just like free fall.

vx

COMPONENTS OF MOTION

Horizontal component¡ Horizontal velocity = constant¡ No gravitational force acts horizontally

Vertical component¡ Vertical velocity = accelerated¡ Object is in free fall with a constant

downward acceleration due to gravity

¡ Result = curved path

In general, any 2D motion is made up of two, independent, simultaneous one-dimensional motions at right angles to each other

INDEPENDENCE OF MOTION IN 2D¡ Motions in the x and y directions are independent of each other¡ ax = 0 (constant acceleration)¡ ay = - g = - 9.8 m/s2

(free-fall; acceleration due to gravity)

STRATEGIES FOR SOLVING PROJECTILE MOTION

¡ Resolve vectors into components¡ Time is the Key! Vertical & horizontal

motion are connected through the variable time

¡ Solving time in one of the dimensions (vertical or horizontal) automatically gives you the time for the other dimension

¡ List all known and unknown variables –helpful to make a table with information for horizontal and vertical components of motion

Vertical Horizontal

vi

vf

a

t

d

PROJECTILES LAUNCHED HORIZONTALLY

¡ A projectile launched horizontally has no initial vertical velocity

¡ Vertical motion is identical to that of a dropped object

¡ Downward velocity increases regularly because of the acceleration due to gravity

DISPLACEMENT CALCULATIONS FOR HORIZONTAL PROJECTION:

For any constant acceleration: d = vit + ½ at2

Horizontal displacement: dx = vixt

Vertical displacement: dy = ½ at2

For the special case of horizontal projection:ax = 0 ; ay = g ; viy = 0 ; vix = vfx

VELOCITY CALCULATIONS FOR HORIZONTAL PROJECTION:

For any constant acceleration: d = vit + ½ at2

Horizontal velocity: vx = vix

Vertical velocity: vy = vi + at

For the special case of a projectile:ax = 0 ; ay = g ; viy = 0 ; vix = vfx

SAMPLE PROBLEMA rock is horizontally launched at 15 m/s off from a cliff 140 m high. How far from the base of the cliff did the rock land?

Vertical Horizontal

vi

vf

atd

0 m/s 15 m/s15 m/s

- 140 m

- 9.8 m/s2 0 m/s2

??

1. List all known and unknown variables2. Common variable between vertical &

horizontal components = time§ Use d = vit + ½ at2§ - 140 m = 0 + (1/2 * -9.8 m/s2 * t2)§ t = 5.34 s

3. Find range (how far from base) § Use dx = vixt§ dx = (15 m/s)(5.34 s)§ dx = 80.1 m

t = 5.34 s80.1 m

YOU TRY!Given the following situation of a marble in motion on a rail (neglect friction), answer the following:

a) Once the ball leaves the table, calculate how long it will take for the ball to hit the floor.0.553 s

b) How far will the ball travel horizontally before hitting the floor?5.53 m

*Impact Velocity:How fast is it moving the instant before it hits the ground?¡ Velocity is a vector quantity; find the two components, then the

magnitude, or speed. Use the Pythagorean relationship to find v.

¡ vfx = 10 m/s¡ vfy = viy + at

= 0 + (-9.8 m/s2 * 0.553 s) = - 5.4 m/s¡ v = vx

2 + vy2

= (10 m/s)2 + (− 5.4 m/s)2

= 11.4 m/s

PROJECTILES LAUNCHED AT AN ANGLE¡ When a projectile is launched at an angle, its initial velocity has an X-

component (vx) and a Y-component (vy)¡ First resolve initial velocity (vi) into components:

¡ To find the components of final position and velocity:

viq vix = vi cosθ ; viy = vi sinθ

Displacement:dx = vixtdy = viyt + ½ at2

Velocity:vfx = vix

vfy = viy + at

TRAJECTORY OF PROJECTILE

x

y

Range

Range: how far a projectile travels horizontally

TRAJECTORY OF PROJECTILE

x

y

¡ Maximum height: the height of the projectile when the vertical velocity is zero and the projectile has only its horizontal velocity component

MaximumHeight

Vy = 0

SYMMETRY IN PROJECTILE MOTION

qvo q-

vo

SAMPLE PROBLEMA soccer ball is kicked at a 25° angle with the ground at 20 m/s.¡ How far did it travel along the ground?

Vertical Horizontal

vi

vf

atd

1. List all known and unknown variables2. Resolve vertical & horizontal components3. Common variable between vertical &

horizontal components = time§ Use vfy = viy + at§ t = 1.7 s

4. Find range (how far from base) § Use dx = vixt§ dx = (18 m/s)(1.7 s) = 31 m

- 9.8 m/s2 0 m/s2

20sin25 20cos25

0 ?

8.5 m/s 18 m/s18 m/s-8.5 m/s

?

due to horizontal velocity of projectile is constant

due to the symmetrical nature of a projectile’s trajectory

1.7 s31 m

Maximum Height: How high did the ball travel?¡ vf

2 = vi2 + 2ad

¡ d = (vf2 "vi2)

2a

¡ = (02 " 8.52)2(−9.8)

¡ d = 3.7 m

SAMPLE PROBLEM (CONT.)

Recall: At max height, Vy = 0

YOU TRY!

You kick a soccer ball at an angle of 40° above the ground with a velocity of 20 m/s.¡ How far away from you will it hit the ground?

¡ 40.2 m¡ What is the max height the ball will go?

¡ 8.5 m