Download - Projectile Motion
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Projectile Motion
YouTube - Baxter NOOOOOOOOOO
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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.
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Amazing facts!
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Amazing facts!
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Amazing facts!
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Amazing facts!
Mr Porter can demonstrate this for you.
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Amazing facts!
Why?
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Vertical and horizontal
Their vertical motion can be considered separate from their horizontal motion.
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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.
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Watch that dog!
Imagine a dog being kicked horizontally off the top of a cliff (with an initial velocity vh).
vh
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Parabola
Assuming that there is negligible air resistance, he falls in the path of a parabola.
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Parabola
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Parabola
Why?
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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
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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
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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
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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!
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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
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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
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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
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Parabola
12.4 metres
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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
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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
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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
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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
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Let’s try some questions.
Page 139 Questions 1, 2, 3 and 4.
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Starting with non-horizontal motion
Woof! (help)
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Starting with non-horizontal motion
30°
25 m.s-1
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Starting with non-horizontal motion
1. Split the initial velocity into vertical and horizontal components
vh = 25cos30°
vv = 25sin30°
30°
25 m.s-1
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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
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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
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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
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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°
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Starting with non-horizontal motion
6. Don’t forget some problems can also be answered using energy.
30°
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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
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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°
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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°
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Let’s try some harder questions.
Page 140 Questions 10, 11,
12, 19.
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Investigation