Projectile Motion

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# Projectile Motion - PowerPoint PPT Presentation

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!.

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Presentation Transcript
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!

Mr Porter can demonstrate this for you.

Vertical and horizontal

Their vertical motion can be considered separate from their horizontal 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.

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.

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/vh2which you may (!) recognise as the formula of a parabola.

Another piece of ultra cool physics!

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 mThe 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

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

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.

energy = ½mv2

V = ?

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

energy at top = energy at bottom

306.5m = 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

Let’s try some questions.

Page 139 Questions 1, 2, 3 and 4.

Starting with non-horizontal motion
• Split the initial velocity into vertical and horizontal components

vh = 25cos30°

vv = 25sin30°

25 m.s-1

30°

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

25 m.s-1

30°

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

25 m.s-1

30°

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

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°

Starting with non-horizontal motion

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

30°

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

25 m.s-1

30°

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°

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°

Let’s try some harder questions.

Page 140 Questions 10, 11, 12, 19.