Aim: How can we solve projectile motion (trajectory) problems?

Do Now:

An object is thrown straight up with a velocity of 10 m/s. How long is it in the air for?

vi = 10 m/s

a = -9.8 m/s2

d = 0 m

t = ?

dy = viyt + ½ayt2

t = 2 s

ProjectilesProjectiles

Launched Launched at an at an angleangle

M M

http://www.youtube.com/watch?v=VgqE87UmQ10

Evil Knievel

Water Slide

Projectiles have both x and y Projectiles have both x and y componentscomponents

Horizontal motion:

Velocity is constant (the same as horizontally fired objects)

Vertical Motion

The same as vertically fired objects

Horizontal motion – look at the dots

Vertical motion – look at the dots

Physlet Physics

Ch. 3

I.3.4

Initial VelocityInitial Velocity

Unlike horizontal projectiles, there are both Unlike horizontal projectiles, there are both vvix ix and vand viyiy

Initial velocity can be resolved into x and y Initial velocity can be resolved into x and y componentscomponents

vi

vix = vicosθ

viy= visinθ

θ

Can’t remember your trig?!Can’t remember your trig?!

It’s in your reference table!!It’s in your reference table!!

Is there an easier way to remember Is there an easier way to remember if vif vix ix or vor viyiy is sin is sinθθ or cos or cosθθ??

It’s in your reference table!!It’s in your reference table!!

In this case, A = vi

Now what?Now what?

Since you can determine your vSince you can determine your v ixix and v and viyiy, the rest , the rest we can figure out!!!we can figure out!!!Horizontal velocity is still constantHorizontal velocity is still constantHorizontal acceleration is still zeroHorizontal acceleration is still zeroVertical acceleration is -9.81 m/sVertical acceleration is -9.81 m/s2 2 (just like in (just like in vertically fired objects)vertically fired objects)Vertical vVertical vff at max height is 0 m/s (just like in at max height is 0 m/s (just like in vertically fired objects)vertically fired objects)Vertical displacement is 0 m (the object starts on Vertical displacement is 0 m (the object starts on the ground and ends on the ground)the ground and ends on the ground)Time to max height = ½ total timeTime to max height = ½ total time

ExampleYou kick a soccer ball into the air at a 30o angle with a velocity of 30 m/s. Find the time in the air.

First step: Find vix and viy

x y

a = -9.8 m/s2

d = 0 m

t = ?

t = ?

vix = vicosθ

vix = (30 m/s)cos30

vix = 26 m/s

viy = visinθ

viy = (30 m/s)sin30

viy = 15 m/s

x y

a = -9.8 m/s2

d = 0 m

t = ?

viy = 15 m/s

t = ?

vix = 26 m/s

d = vit + ½at2

0 m = (15 m/s)t + ½(-9.8 m/s2)t2

0 = 15t – 4.9t2

4.9t2 = 15t

4.9t = 15

t = 3.1 s

Find the maximum height x y

a = -9.8 m/s2

d = 0 m

viy = 15 m/s

t = 3.1 s

dmax = ?

vfmax = 0 m/s

tmax = 1.55 s

vix = 26 m/s

t = 3.1 s

Choose a formula:

vfmax2 = viy

2 + 2admax

dmax = viytmax + ½atmax2

dmax = viytmax + ½atmax2

dmax = (15 m/s)(1.55 s) + ½(-9.8 m/s2)(1.55 s)2

dmax = 11.48 m

Solve for the range

x y

a = -9.8 m/s2

d = 0 m

viy = 15 m/s

t = 3.1 s

dmax = ?

vfmax = 0 m/s

tmax = 1.55 s

vix = 26 m/s

t = 3.1 s

d = ?

Remember:

vix = constant

Therefore, vix = v

t

dv

s

dsm

1.3/26

d = 80.6 m

Thinking questionThinking question

What angle gives the What angle gives the greatest range?greatest range?

45o

Which angle gives the: greatest range?

greatest height?

45°

90°

Another example problemAnother example problem

A projectile is fired at an angle of 22A projectile is fired at an angle of 22o o with an initial velocity of 120 meters/second.

How highwill the projectile travel? How far will the projectile go?

Pages 130-131Pages 130-131