*Moving in the x and y direction

*A projectile is an object shot

through the air. This occurs in a

parabola curve.

Object

droppedObject

thrown up

Object thrown at an

angle

projectile- any object that moves through the air

or through space, acted on only by gravity

(and air resistance, if any)

The vertical acceleration of a

projectile is caused by gravity, so

ay = -9.8 m/s2

Parabolic

Trajectory

*g remains constant (g= -9.8m/s2)

*a in the x direction is 0 because gravity is not

acting on it.

*Neglect air resistance

*Neglect the effects of the earths rotation

Projectiles launched horizontally

To find how far the ball falls, you

use the formula. y =viyt + 1/2gt2

1st second- 5m

After 2 seconds- 20m

After 3 seconds- 45m

The curved path of a

projectile produced is a

parabola (caused by both

horizontal motion and vertical

motion. It must accelerate

only in the vertical direction)

*

*The projectile will experience two:

*Accelerations (ax= o and aY= -9.8m/s2)

*Velocities

*Displacements

Upwardly Launched Projectiles

When a projectile is launched at an upward

angle, it follows a curved path and finally

hits the ground because of gravity.

The Vertical distance a cannonball falls below “imaginary

path if no gravity” is the same vertical distance it

would fall if it were dropped from rest & had been

falling for the same amount of time.

*Draw a free body diagram with a coordinate

system.

*Divide the information into x and y components

*Look at your formulas and decided which

one(s) to use.

Formulas for horizontal and vertical

motion of a projectile

Objects that have been thrown will have a

horizontal velocity that stays the same (no

horizontal acceleration ax = 0m/s2)

So vfx =vix in the second formula and third

formulas under horizontal motion.

(X) Horizontal (Y) Vertical

xf-xi = vixt + ½ axt 2 yf-yi = viyt + ½ ayt

2

vfx = vix + axt vfy = viy + ayt

vfx2 - vix

2 = 2ax(xf-xi) vfy2 = viy

2 + 2ay (yf-yi)

*

This equation only works when y and y0 are both the same magnitude

𝑡 =2𝑦

𝑔

y0 y

If a ball is thrown up in the air from a moving truck, where will it land?

(Ignore air resistance)

In front of the truck, behind the truck, or back in the truck

Where will a package land if it is released

from a plane?

Behind the plane, in front of the plane

below the plane

What is the horizontal distance covered by

an arrow that was shot through the air at a

600 angle with a velocity of 55 m/s?

Given

v = 55m/s

vx=27.5 m/s

vyo=47.6m/s

ax=0

ay=-9.8m/s2

t=?

x =?

Solve

Vx = cos 60(55m/s)=27.5 m/s

Vyo = sin60(55m/s)=47.6m/s

vy=vyo +ayt

x = Vx t (we need time)

600

55m/s y

Vx x

0 = 47.6m/s + -9.8m/s2 t

-47.6m/s = -9.8m/st

4.86 s =t

x = 27.5 m/s(9.7s)

x = 266.75m

Total time in the air 4.86s x 2 = 9.7s

Need to find time first!

To find x dist: x = vx t

* A boat heading due north

crosses a river with a

speed of 10.0 km/h. The

water in the river has a

speed of 5.0 km/h due

east.

Moving frame of reference

In general we have PA PB BAv v v

(a) Determine the velocity of the boat.

(b) If the river is 3.0 km wide how long does it take to cross it?