Projectile Motion

39
Projectile Motion YouTube - Baxter NOOOOOOOOOO

description

Projectile Motion. YouTube - Baxter NOOOOOOOOOO. 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!. - PowerPoint PPT Presentation

Transcript of Projectile Motion

Page 2: 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.

Page 3: Projectile Motion

Amazing facts!

Page 4: Projectile Motion

Amazing facts!

Page 5: Projectile Motion

Amazing facts!

Page 6: Projectile Motion

Amazing facts!

Mr Porter can demonstrate this for you.

Page 7: Projectile Motion

Amazing facts!

Why?

Page 8: Projectile Motion

Vertical and horizontal

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

Page 9: Projectile Motion

Vertical and horizontal

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.

Page 10: Projectile Motion

Watch that dog!

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

vh

Page 11: Projectile Motion

Parabola

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

Page 12: Projectile Motion

Parabola

Page 13: Projectile Motion

Parabola

Why?

Page 14: Projectile Motion

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

Page 15: Projectile Motion

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

Page 16: Projectile Motion

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

Page 17: Projectile Motion

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!

Page 18: Projectile Motion

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

Page 19: Projectile Motion

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

Page 20: Projectile Motion

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

Page 21: Projectile Motion

Parabola

12.4 metres

Page 22: Projectile Motion

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

Page 23: Projectile Motion

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

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

Page 24: Projectile Motion

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

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

V = ?

energy = ½mv2

Page 25: Projectile Motion

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

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

Page 26: Projectile Motion

Let’s try some questions.

Page 139 Questions 1, 2, 3 and 4.

Page 27: Projectile Motion

Starting with non-horizontal motion

Woof! (help)

Page 28: Projectile Motion

Starting with non-horizontal motion

30°

25 m.s-1

Page 29: Projectile Motion

Starting with non-horizontal motion

1. Split the initial velocity into vertical and horizontal components

vh = 25cos30°

vv = 25sin30°

30°

25 m.s-1

Page 30: Projectile Motion

Starting with non-horizontal motion

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

Page 31: Projectile Motion

Starting with non-horizontal motion

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

Page 32: Projectile Motion

Starting with non-horizontal motion

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

Page 33: Projectile Motion

Starting with non-horizontal motion

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°

Page 34: Projectile Motion

Starting with non-horizontal motion

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

30°

Page 35: Projectile Motion

Starting with non-horizontal motion

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

Page 36: Projectile Motion

Starting with non-horizontal motion

6. At the highest point,

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

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

30°

Page 37: Projectile Motion

Starting with non-horizontal motion

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°

Page 38: Projectile Motion

Let’s try some harder questions.

Page 140 Questions 10, 11,

12, 19.

Page 39: Projectile Motion

Investigation