Projectile Motion YouTube - Baxter NOOOOOOOOOO. Projectile Motion 9.1Projectile motion (AHL) 9.1.1State the independence of the vertical and the horizontal

Projectile Motion

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Projectile Motion

9.1 Projectile motion (AHL) 9.1.1 State the independence of the vertical and the horizontal

components of velocity for a projectile in a uniform field.

9.1.2 Describe and sketch the trajectory of projectile motion as parabolic in the absence of air resistance.

9.1.3 Describe qualitatively the effect of air resistance on the trajectory of a projectile.

9.1.4 Solve problems on projectile motion.

Amazing facts!

If a gun is fired horizontally, and at the same time a bullet is dropped from the same height. They both hit the ground at the same time.

Amazing facts!

Amazing facts!

Amazing facts!

Amazing facts!

Amazing facts!

Why?

Vertical and horizontal

Their vertical motion can be considered separate from their horizontal motion.

Vertical and horizontal independent of each other

Vertically, they both have zero initial velocity and accelerate downwards at 9.8 m.s-2. The time to fall the same vertical distance is therefore the same.

Horizontally velocity remains constant

Projectiles (Half Trajectory)

An object projected sideways through the air will follow a curved trajectory.

horizontal motion (steady speed)

vertical motion

The horizontal and vertical motions should be treated separately.

Time is the only quantity common to both.

accelerates downwards at -9.8 ms-

2

tD

V HH

At any point in its trajectory, the velocity of a projectile has two components.

• one vertical, VV

• the other horizontal, VH

The resultant velocity is found drawing a vector diagram and add the vectors together, TIP to TAIL.

Vector Diagram

horizontal velocity

vertical velocity

resultant/actual

velocity

vh

vv

This is an example of a ‘half-trajectory.’

GREEN – actual motion

RED – vertical motion

BLUE – horizontal motion

Watch that dog!

Imagine a dog being kicked horizontally off the top of a cliff (with an initial velocity vh).

vh

Parabola

Assuming that there is negligible air resistance, he falls in the path of a parabola.

Parabola

Parabola

Why?

Why a parabola?

We can consider his motion to be the sum of his horizontal motion and vertical motion.

We can treat these separately

vh

Horizontal motion

Assuming no air resistance, there are no horizontal forces.

This means horizontally

the dog moves with

constant speed vh

vh

Horizontal distance travelled (x) = vht

Vertical motion

Assuming no air resistance, there is constant force downwards (=mg).

This means vertically the

dog moves with constant

acceleration g = 9.8 m.s-2

Vertical distance travelled (y) = uvt + ½gt2

Parabolic motion

Since y = ½gt2 (if u = 0) and x = vht,

y = ½gx2/vh2 which you may (!) recognise as

the formula of a parabola.

Another piece of ultra cool physics!

30 ms-1

Example

A ball is kicked horizontally off an embankment, with a velocity of 30 ms-1.

It lands 24 m from the base of the embankment.

(a) Calculate how long the ball was in flight.

24 m

tD

V HH

t24

30

3024

t

s 0.8t common to

horizontal and vertical motions

Horizontal Vertical

-1ms 30u

m 24s

-2ms -9.8a

(b) Calculate the horizontal velocity just before hitting the ground.

s 0.8t

s 0.8t

-2ms 0a

-1ms 0u

travels horizontally at steady speed

– no acceleration horizontally

not initially falling down, so speed of

zero in vertical direction

acted upon by gravity

Horizontal

atuv

0.8030 -1ms 30v

(c) Calculate the vertical velocity just before hitting the ground.

Vertical

atuv 0.89.8-0

-1ms -7.84v

(d) How high is the embankment?

Vertical

-2ms -9.8as 0.8t

-1ms 0u-1ms -7.84v

means 7.84 ms-1 downwards

2at21

uts

20.89.8-21

0.80

m 3.14s

so height of the embankment is 3.14 m

means ball fell through distance of

3.14 m

(e) Calculate the resultant velocity of the ball, just before hitting the ground.

30 ms-

1

-7.8 ms-1

velocity

θ

Size

By Pythagoras:

222 cba

222 7.84-30velocity resultant 61.5900

961.5velocity resultant -1ms 31

Direction

hyp

adjθ cos

3130

θ cos

0.97cosθ 1

14.6θ

horizon below 14.6 of angle at ms 31velocity resultant -1

30 ms-

1

-7.8 ms-1

velocity

θ

Q1. A ball is kicked off a cliff with a horizontal speed of 16 ms-

1.

The ball hits the ground 2.2 s later.

(a) Calculate the height of the cliff.

(b) Calculate the distance between the foot of the cliff and where the ball lands.

(c) Calculate the vertical component of the balls velocity just before it hits the ground.

(d) Calculate the balls velocity as it hits the ground.

23.7 m

35.2 m

21.6 ms-1

26.9 ms-1 at angle of 53.5° below

horizon

You may want to draw a diagram to help you get started !!!

Q2. A ball is kicked off a cliff with a horizontal speed of 22 ms-1. the ball hits the ground 1.5 s later.

(a) Calculate the height of the cliff.

(b) Calculate the horizontal distance from the foot of the cliff, to where the ball lands.

(c) Calculate the vertical component of the balls velocity as it hits the ground.

(d) Calculate the balls actual velocity as it hits the ground.

11 m

14.7 ms-1

26.5 ms-1 at angle of 34° below

horizon

33 m

You may want to draw a diagram to help you get started !!!

Example

A dog is kicked off the top of a cliff with an initial horizontal velocity of 5 m.s-1. If the cliff is 30 m high, how far from the cliff bottom will the dog hit the ground?

5 m.s-1

30 m

Example

Looking at vertical motion first:

u = 0, a = 9.8 m.s-2, s = 30 m, t = ?

s = ut + ½at2

30 = ½ x 9.8 x t2

t2 = 6.1

t = 2.47 s

The dog hits the ground after 2.47 seconds (yes!)

5 m.s-1

30 m

Example

Now look at horizontal motion:

Constant speed (horizontally) = 5 m.s-1

Time of fall = 2.47 seconds

Horizontal distance travelled = speed x time

Horizontal distance travelled = 5 x 2.47

= 12.4

m The dog hits the ground 12.4 metres from the base of the cliff

5 m.s-1

30 m

Parabola

12.4 metres

What is the dog’s speed as he hits the ground?

To answer this it is easier to think in terms of the dog’s total energy (kinetic and potential)

5 m.s-1

30 m


Total energy at top = ½mv2 + mgh

Total energy = ½m(5)2 + mx9.8x30

Total energy = 12.5m + 294m = 306.5m

5 m.s-1

30 m


At the bottom, all the potential energy has been converted to kinetic energy. All the dog’s energy is now kinetic.

V = ?

energy = ½mv2


energy at top = energy at bottom306.5m = ½mv2

306.5 = ½v2

613 = v2

V = 24.8 m.s-1

(Note that this is the dog’s

speed as it hits the ground,

not its velocity.

v = 24.8 m.s-1

Projectiles (Full Trajectory)

A projectile does not need to be an object falling, but could be an object fired at angle to the horizontal.

The subsequent motion would bemax height

θ

If air resistance is ignored, the trajectory has an axis of symmetry about the mid point (maximum height).

So the time taken to reach the maximum height is the same as the time taken to fall back to the ground.

Various calculations can be made, but firstly, the initial velocity must be split into its horizontal and vertical components.

Horizontal

a = 0 ms-2

Vertical

a = -9.8 ms-2

Starting with non-horizontal motion

Woof! (help)


30°

25 m.s-1


1. Split the initial velocity into vertical and horizontal components

vh = 25cos30°

vv = 25sin30°

30°

25 m.s-1


2. Looking at the vertical motion, when the dog hits the floor, displacement = 0

Initial vertical velocity = vv = 25sin30°

Acceleration = - 9.8 m.s-2

30°

25 m.s-1


3. Using s = ut + ½at2

0 = 25sin30°t + ½(-9.8)t2

0 = 12.5t - 4.75t2

0 = 12.5 – 4.75t

4.75t = 12.5

t = 12.5/4.75 = 2.63 s

30°

25 m.s-1


4. Looking at horizontal motion

Ball in flight for t = 2.63 s travelling with constant horizontal speed of

vh = 25cos30° = 21.7 m.s-1.

Distance travelled = vht = 21.7x2.63 = 57.1m

30° 57.1m


5. Finding maximum height? Vertically;

v = 0, u = 25sin30°, t = 2.63/2

s = (u + v)t = 12.5x1.315 = 8.2m

2 2

30°


6. Don’t forget some problems can also be answered using energy.

30°


6. Don’t forget some problems can also be answered using energy.

As dog is fired total energy = ½m(25)2

30°

25 m.s-1


6. At the highest point,

total energy = KE + GPE =½m(25cos30°)2 + mgh

As dog is fired total energy = ½m(25)2

30°


6. So ½m(25cos30°)2 + mgh = ½m(25)2

½(21.65)2 + 9.8h = ½(25)2

234.4 + 9.8h = 312.5

9.8h = 78.1

h = 8.0 m

30°

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Projectile Motion YouTube - Baxter NOOOOOOOOOO. Projectile Motion 9.1Projectile motion (AHL) 9.1.1State the independence of the vertical and the horizontal