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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!.

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## Projectile Motion

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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!**Mr Porter can demonstrate this for you.**Amazing facts!**Why?**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.**Parabola**Why?**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**Woof! (help)**Starting with non-horizontal motion**25 m.s-1 30°**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.

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