Download Projctiles - Hinchingbrooke

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.
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
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
x
Maximum height
Assuming O is at ground level, a projectile reaches maximum height,
H, when its vertical component of velocity is 0.
Using v2 = u2 + 2as ,
u = U sin θ
U
v=0
θ
a = –g
O
s=H
So,
02 = U2sin2θ – 2gH
2gH = U2sin2θ
U 2sin2
H=
2g
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
0 = Usinθ – gt
So,
The time to reach maximum height is therefore

2U sin
The time of flight =
g
U sin
.
g
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:
2U sin
t=
g
The range of the flight, R, is given by the horizontal velocity × the
time of flight.
So,
U cos ×2U sin
R=
g
2U 2sin cos
R=
g
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:
g
2
y  x tan   2
x
2u cos 2 
y
U
θ
O
x
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?
g
2
y  x tan   2
x
2u cos 2 
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. 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 36.9o 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.
a) 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.
b) Eliminate t between the two equations to show that the trajectory
equation is:
3x x 2
y
4

320
c) How high will the ball be when it has travelled 10m horizontally?
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.