Lesson Objective: Understand that we can use the suvat equations in 2d by considering the two directions independently Be able to solve problem projectile problems where items are launched horizontally in 2d

Projectiles. What are they? You tube – myth buster – bullet You tube – MIT monkey shoot For all our projectile questions we assume: Point particles – no spinning to provide lift! No air resistance - so that any acceleration is constant

Main principle: An objects movement in a vertical direction and a horizontal direction are independent: Acceleration acting downwards (gravity) does not affect the horizontal motion. Similarly acceleration in the horizontal plane does not affect the vertical motion.

A stone is thrown horizontally from the top of a cliff at 10ms-1 A stone is thrown horizontally from the top of a cliff at 10ms-1. The cliff is 80m high above the sea. Find the time taken for the stone to reach the sea, and its distance from the base of the cliff when it hits the water. Find the distance of the stone in example 1 from its starting position after 2 seconds. Find also the magnitude and direction of the stone’s velocity at this time.

Lesson Objective Projectiles launched at an angle to the horizontal from flat ground

A bomb is to be dropped from an aeroplane which is flying steadily at 1000m with a speed of 200ms-1. How far (horizontally) should the plane be from the target before it releases the bomb, and how long will the bomb take to hit the target?

A stone is thrown at 10ms-1 at an angle of 30° to the horizontal from flat level ground. Find: a) the time for which the stone is airborne. b) the ‘range’ of the stone (i.e. how far it is from its starting point when it lands). c) the speed and direction of the stone after 0.3 seconds.

Maximum height, range and time in flight Suppose a particle is initially launched from O with a velocity U and at an angle θ to the horizontal. y U Maximum height θ O x Range The main things that can be calculated are: The maximum height reached by the particle The total time that the particle is in flight The range of the flight

Maximum height Assuming O is at ground level, a projectile reaches maximum height, H, when its vertical component of velocity is 0. U θ O Using v2 = u2 + 2as , u = U sin θ v = 0 a = –g s = H So, 02 = U2sin2θ – 2gH The projectile reaches maximum height at the point when it stops moving upwards and starts moving downwards – that is, when the vertical component of velocity is 0. This formula need not be learnt since students are expected to be able to derive it in an exam situation. 2gH = U2sin2θ

Time of flight The path of a projectile is symmetrical on horizontal ground. This means that if we find the time taken to reach the maximum height, the total time that the particle is in flight will be double that amount. Using v = u + at , u = U sin θ v = 0 a = –g So, 0 = Usinθ – gt Again, this formula should not be learnt but students need to be able to derive it. The time to reach maximum height is therefore . The time of flight = 

Range of flight We have just shown that the time of flight for a particle launched from O with a velocity U and at an angle θ to the horizontal is: t = The range of the flight, R, is given by the horizontal velocity × the time of flight. So, When considering the horizontal distance travelled there is no acceleration in this direction. Hence, the horizontal distance travelled can be found using horizontal velocity × time. If required, the formula given here for the range can be further simplified using the identity sin2θ = 2sinθcosθ, although knowledge of this identity is not expected at this stage. Again, the derivation and not the formula should be learnt.

A Ming vase is thrown from the top of a tower 15m high, at a speed of 10ms-1 and at an angle of 75° to the downward vertical. Find a) the time taken for the vase to smash on the ground. b) its horizontal range. c) the speed and direction of motion of the vase just before impact.

The Trajectory Equation A projectile is launched from a height of 0m with an initial speed of 20ms-1 at an angle of 30 degrees to the horizontal. What will be the height of the projectile when it has travelled 3m horizontally?

The Trajectory Equation: θ O x y This is an equation that relates the horizontal displacement, x, to the vertical displacement, y. It is great for solving problems like the last one, where you know the ‘position’/displacement of an object at a particular time.

A projectile is launched from a height of 0m with an initial speed of 20ms-1 at an angle of 30 degrees to the horizontal. What will be the height of the ball when it has travelled 3m horizontally?

Proof of the trajectory equation: Consider displacement in both directions independently for a specified time, t. Eliminate ‘t’ from the vertical displacement equation.

An archer sits on top of a 10m tower, and aims directly at a rabbit 20m from the base of the tower. If the arrow is fired at 28ms-1, by how much does it fall short of the rabbit?

LESSON OBJECTIVE Consolidate work on 2d kinematics: projectiles Practise some exam questions

LESSON OBJECTIVE Understand what we mean by a trajectory equation Be able to find the trajectory equation and use it to solve projectile problems Consider volley ball player serving a ball. They serve it at a height of 1m, 5m away from the central net which is 2m high. If they serve the ball at 8ms-1 at an angle of 30 degrees to the horizontal will the ball clear the Net?

Consider volley ball player serving a ball Consider volley ball player serving a ball. They serve it at a height of 1m, 5m away from the central Net which is 2m high. If they serve the ball at 8ms-1 at an angle of 30 degrees to the horizontal will the ball clear the Net? y axis x axis

The trajectory equation is an equation that relates the vertical displacement directly (y) to the horizontal displacement (x). To find the trajectory equation Write down the vertical displacement (y) of the object in terms of ‘t’ Write down the horizontal displacement (x) of the object in terms of ‘t’ Eliminate ‘t’ to relate y directly to x. The trajectory equation is great when you need to find out if a an item passes through a particular point/co-ordinate, or to find the height of a an object directly from its horizontal position.

A particle is projected with initial velocity 50ms-1 at an angle of 36 A particle is projected with initial velocity 50ms-1 at an angle of 36.9o to the horizontal. The point of projection is taken to be the origin, with the x axis horizontal and the y axis vertical in the plane of the particle’s motion. Show that at time t, the height, y, of the particle in metres is given by y = 30t – 5t2 and write down the corresponding expression for x. Eliminate t between the two equations to show that the trajectory equation is: How high will the ball be when it has travelled 10m horizontally?

A batsman hits a cricket ball 0. 5 metres above the ground A batsman hits a cricket ball 0.5 metres above the ground. The ball leaves the bat with a velocity of 30ms-1, and is caught by a fielder 45 metres away, 1.5 metres above the ground. Find the two possible angles to the horizontal at which the ball leaves the bat.

Lesson Objective Projectiles Consolidation TASKS Answer Part 1 of the Phydeaux problem (Starter) Answer Part 2 of the Phydeaux problem (you may choose which one) Draw a nice poster for me illustrating this problem with a neat and correct working solution to the parts you have answered http://www.youtube.com/watch?v=0ImrvABD9AU

Phydeaux the performing dog is to be fired by a cannon across the 200m wide River Danube. The muzzle velocity is 45ms-1 and the angle of elevation of the barrel is 35°. Does Phydeaux reach the other side, or does he get wet?

Understandably dissatisfied with his manager, Phydeaux gives him the sack, and considers improving his act in three ways: a) increasing the velocity of projection u. b) changing the angle of elevation θ of the barrel. c) placing the cannon on a platform of height h. In each case all the other variables are unchanged. Find the least values of u, θ and h to enable him to reach the other side.