Projectile Motion

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    02-Jan-2016
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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

  • 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