# Projectile Motion

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02-Jan-2016Category

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### Transcript of Projectile Motion

Projectile MotionYouTube - 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!

Mr Porter can demonstrate this for you.

Amazing facts!

Why?

Vertical and horizontalTheir vertical motion can be considered separate from their horizontal motion.

Vertical and horizontalVertically, 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

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

Parabola

ParabolaWhy?

Why a parabola?We can consider his motion to be the sum of his horizontal motion and vertical motion.

We can treat these separatelyvh

Horizontal motionAssuming no air resistance, there are no horizontal forces.

This means horizontally the dog moves with constant speed vhvhHorizontal distance travelled (x) = vht

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

This means vertically the dog moves with constant acceleration g = 9.8 m.s-2Vertical distance travelled (y) = uvt + gt2

Parabolic motionSince y = gt2 (if u = 0) and x = vht, y = gx2/vh2 which you may (!) recognise as the formula of a parabola.

Another piece of ultra cool physics!

ExampleA 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-130 m

ExampleLooking at vertical motion first:

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

s = ut + at2 30 = x 9.8 x t2t2 = 6.1t = 2.47 sThe dog hits the ground after 2.47 seconds (yes!)5 m.s-130 m

ExampleNow look at horizontal motion:

Constant speed (horizontally) = 5 m.s-1Time 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 cliff5 m.s-130 m

Parabola12.4 metres

What is the dogs speed as he hits the ground?To answer this it is easier to think in terms of the dogs total energy (kinetic and potential)5 m.s-130 m

What is the dogs speed as he hits the ground?Total energy at top = mv2 + mgh

Total energy = m(5)2 + mx9.8x30Total energy = 12.5m + 294m = 306.5m5 m.s-130 m

What is the dogs speed as he hits the ground?At the bottom, all the potential energy has been converted to kinetic energy. All the dogs energy is now kinetic.V = ?energy = mv2

What is the dogs speed as he hits the ground?energy at top = energy at bottom306.5m = mv2306.5 = v2613 = v2V = 24.8 m.s-1(Note that this is the dogs speed as it hits the ground, not its velocity.v = 24.8 m.s-1

Lets try some questions.

Page 139 Questions 1, 2, 3 and 4.

Starting with non-horizontal motionWoof! (help)

Starting with non-horizontal motion3025 m.s-1

Starting with non-horizontal motionSplit the initial velocity into vertical and horizontal components

vh = 25cos30vv = 25sin303025 m.s-1

Starting with non-horizontal motion2. Looking at the vertical motion, when the dog hits the floor, displacement = 0Initial vertical velocity = vv = 25sin30Acceleration = - 9.8 m.s-2

3025 m.s-1

Starting with non-horizontal motion3. Using s = ut + at20 = 25sin30t + (-9.8)t20 = 12.5t - 4.75t20 = 12.5 4.75t4.75t = 12.5t = 12.5/4.75 = 2.63 s

3025 m.s-1

Starting with non-horizontal motion4. Looking at horizontal motionBall 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

3057.1m

Starting with non-horizontal motion5. Finding maximum height? Vertically;v = 0, u = 25sin30, t = 2.63/2s = (u + v)t = 12.5x1.315 = 8.2m2 2

30

Starting with non-horizontal motion6. Dont forget some problems can also be answered using energy.

30

Starting with non-horizontal motion6. Dont forget some problems can also be answered using energy.As dog is fired total energy = m(25)2

3025 m.s-1

Starting with non-horizontal motion6. At the highest point, total energy = KE + GPE =m(25cos30)2 + mghAs dog is fired total energy = m(25)2

30

Starting with non-horizontal motion6. So m(25cos30)2 + mgh = m(25)2(21.65)2 + 9.8h = (25)2234.4 + 9.8h = 312.59.8h = 78.1h = 8.0 m

30

Lets try some harder questions.

Page 140 Questions 10, 11, 12, 19.

Investigation