Key Areas covered Projectiles and satellites. Resolving the motion of a projectile with an initial velocity into horizontal and vertical components and their use in calculations. Comparison of projectiles with objects in free-fall.

What we will do today: Quick revision of equations of motion. Investigate projectiles. Relate the two above.

Resolution (Rectangular Components) of a Vector

The horizontal component xh The vertical component xv Any vector, x, can be resolved into two components at right angles to each other. The horizontal component xh The vertical component xv x xv is equivalent to θ xh

sin θ = xv / x xv = x sin θ cos θ = xh / x xh = x cos θ x xv θ xh

Velocity The vertical and horizontal components of a velocity vector, v, are, respectively: vv = v sin θ vh = v cos θ

CfE Specimen Paper: Section B Q1

CfE Specimen Paper (Solution)

The Equations of Motion

The Equations of Motion u – initial velocity at time t = 0 v – final velocity at time t a – acceleration of object t – time to accelerate from u to v s – displacement of object in time t 1) v = u + at 2) s = ut + ½at² 3) v² = u² + 2as

Method for Tackling Problems Write down all the symbols like this: u = v = s = a = t = Fill in all numbers & values given in question. If there are two directions, use this diagram for + and – values: Choose the best equation to suit the problem. Up + Left - Right + Down -

Projectile Motion (using equations of motion)

Projectile Motion When an object is fired horizontally it has both vertical and horizontal motion. eg a cannon firing a cannon ball

Projectile Motion When a projectile is fired it takes a curved path.

Projectile Motion This is due to two things: Horizontal motion – travelling at constant velocity Vertical motion – constant acceleration due to gravity

Projectile Motion These two motions can be represented in graphs: Horizontal Vertical The distance travelled is equal to the area under the graphs. Hor = l x b Ver = ½ l x b

The motion of a projectile consists of two independent parts or components: 1) constant horizontal velocity (accn = 0) 2) constant vertical acceleration (caused by Earth’s gravitational pull). These motions can be treated separately or in combination depending on the circumstances.

Example 1 – Horizontal Projection A projectile is fired horizontally off a cliff as shown: Find: a) the time of flight b) the range c) the vertical velocity just before impact d) the horizontal velocity just before impact e) the actual (resultant) velocity just before impact 5 ms-1 45 m range

Solution a) Vertically, s = ut + ½at² 45 = 0 + 4.9t² t² = 45 / 4.9 = 9.18 t = 3.03 s b) Horizontally, (remember a = 0!!) Range, s = vht = 5 x 3.03 s = 15.2m

c) Vertical velocity, v = u + at = 0 + (9.8 x 3.03) v = 29.7 ms-1 d) Horizontal velocity, v = 5 ms-1 (constant!) e) v² = 5² + 29.7² v = 30.1 ms-1 tan θ = 29.7 / 5, so θ = 80.4º Actual velocity = 30.1 ms-1 at 80.4º below the horizontal or (170) 5 θ v 29.7

Additional note on projectiles Vv = 0 m/s For projectiles fired from a surface to a maximum height, two points must be noted: 1. The (vertical) velocity at the maximum height is 0 ms-1. 2. The time taken to reach the maximum height from a surface, t1, is equal to the time to return from the maximum height back to the surface, t2. i.e. the time taken to reach the maximum height is half of the total time taken. t2 t1

Example 2 2003 Qu: 21

2008 Qu: 2

2004, Qu: 2

2000 Qu: 21

2009 Qu: 21

Satellite Motion

Satellite Motion Satellite motion is an extension of projectile motion. Satellite Motion Earth click here for Newton's thought experiment

Satellites are constantly ‘falling’ towards the Earth in orbit. They remain in orbit and do not move closer to the Earth because they have a great enough horizontal speed. They do not move off into space because there is a force of gravity towards the centre of the Earth.

How can a satellite be accelerating when it is travelling at constant speed? A satellite has constantly changing velocity (because the direction is constantly changing). This means that a satellite can be accelerating when it has a constant speed.

Questions Activity sheets: Projectiles You should now be able to answer all questions in your class jotter

Answers Projectiles 1. (a) 7·8 s (b) 2730 m (c) directly above box 3. (a) (i) and (ii) Teacher Check (b) 24·7 ms−1 at an angle of 52.6º below the horizontal 4. (a) vH = 5·1 ms−1 , vV= 14·1 ms−1 5. (a) (i) and (ii) Teacher Check (b) 50 ms−1 at 36.9º above the horizontal (c) 40 ms−1 , Teacher Check (d) 45.9 m (e) 240 m 6. (a) 20 ms−1 (b) 20.4 m (c) 4·1 s (d) 142 m 7. (a) 8 s (b) 379 m 8. (a) 15·6 ms−1 (b) and (c) Teacher Check 9. Teacher Check 10. Teacher Check 11. Teacher Check 12. 2 s