Download 1999 Solution Q11

2. Projectile Motion
A projectile is a body that has been thrown or projected, and is travelling freely through the air.
While undergoing projectile motion the object is under the constant unbalanced force of gravity.
When we study projectile motion, no consideration is given to the force projecting the body or to
what happens when it lands. Air resistance is considered to be negligible. The projectile will travel
either: vertically or inclined, depending on the initial angle of projection.
Vertically
Inclined
C
C
B D
A E
Ground Level
D
B
E
A
Ground Level
For both the vertical and inclined projectiles:
 the only force acting is the weight, ie. the bodies are in free fall
 acceleration is always 10 ms-2 downward, (including the point C)
 the instantaneous velocity is tangential to the path
 the total energy (KE & PE) is constant
 between any two points KE = - PE
 paths are symmetrical for time eg.  t (A to B) =  t (D to E):
 t (A to C) =  t (C to E)
 paths are symmetrical for speed eg. speed at A = speed at E; speed at B = speed at D.
 for vertical motion vc = O,
for inclined motion vc  0.
Inclined or oblique projections
 the only force acting is vertically down, so the acceleration and change in velocity are vertical.
 horizontally there is no component of force, so constant horizontal velocity.
 Maximum range is when angle of projection is 450
Horizontal projection
When the body is launched horizontally and follows a parabolic path to the ground it is really the
second half of an inclined projection. The time of flight is controlled by the height from which it is
released. The speed of projection will not affect the time 't' that it takes to land. The 'range' of this
projectile is given by the x = vhorizontal × t.
For projectiles thrown horizontally and dropped from rest, the vertical motions are the same. This
can be shown by a multi-flash photograph.
The interval between the lines represents how far
the ball has travelled in a small time interval.
Notice that the distance between the horizontal
lines increases as the ball descends, indicating
that the ball is speeding up. The distance
between successive vertical lines remains
constant, indicating that the ball is travelling with
a constant velocity in the horizontal direction.
If we analyse the motion by using resolution of vectors we get the following:
x
Uhorizontal
y
Uvertical
Vhorizontal
Vvertical
Horizontal:
velocity always = vhorizontal
acceleration = 0
displacement = x = vhorizontal × t
Vertical:
Velocity v = u + gt
acceleration = g
displacement y = ut + ½ gt2
To find the 'total' velocity, add vvertical and vhorizontal using vectors.
If one projectile was fired horizontally, at the same time that another was dropped (from the same
height), then both objects would hit the ground at the same time. This is because both their
vertical motions were identical. (Same distance to fall, initial speed = 0, and acceleration = -g)
Inclined Projectiles.
ground
vo

initial
The vector
representing the
velocity can be resolved into two components
v0
v0sin 

v0cos 
Horizontal:
velocity always = vhorizontal
vhorizontal = v0cos 
acceleration = 0
displacement = x = v0cos  × t
Vertical (on the way up)
velocity changing v = u - gt
Vvertical = v0sin  - gt
acceleration = -g
velocity changing v = u + gt
v = 0 + gt
acceleration = g
Vertical (on the way down)
The displacement at any time of the motion is: y = ut - ½ gt2 (The same on the way up and
down).
Graphs for projectile motion
motion in the 'x' direction
(right is positive)
motion in the 'y' direction
(up is positive)
acceleration
acceleration
velocity
velocity
displacement
displacement
Symmetrical flights
If there is no air resistance, and the projectile starts and ends at the same height, then the range is
given by:
R=
v2 sin 2
g
R is the range, v is the initial speed and  the angle of
projection. Be careful using this formula, because it only works under the conditions specified
above.
Total Energy (TE)
 If air resistance is negligible, then the total energy of a projectile will always remain
constant throughout the flight. TE = KE + PE =





1 2
mv  mgh .
2
At ground level PE = 0, so TE = KE.
As the projectile rises it gains PE, so it must lose KE. At the top of its flight, the PE is
maximum and the KE is minimum. (the KE is not zero, because the projectile still has some
KE due to its horizontal motion).
At any point on the way up or the way down, the TE is constant.
If you know the horizontal component of the velocity, then you can use this to find the
maximum height.
Use the TE at ground level and work out what the PE must be at the top when vvertical = 0,
but vhorizontal = constant.
Projectile problem methodology
1. Draw a fully labelled diagram
2. Treat the motion as two separate motions, for horizontal v0cos  =
3.
4.
5.
6.
7.
8.
R
t
for vertical, acceleration = -g, v = v0sin 
List the data under vertical and horizontal
For vertical motion use, v2 =u2 + 2ax, and v = u + at. (use a = -g)
Usually given information about one direction and asked to find out something about the other
To go from one direction to another, the common link is the time of flight, t.
Remember that it will take the same time to go up as to come down.
Label the direction (+ or -) for all variables except time.
Examples
2000
Su and Harry go for a day’s surfing with surfboards strapped to the roof of their car. They enter the cliff-top
car-park and brake suddenly as they park their car. A surfboard breaks free from the roof strap, and
continues forward in a horizontal direction with an initial speed of 10 m s -1.
When it breaks free the surfboard is initially 80 m above the surface of the sea.
2000 Question 4
Use one or more of Newton’s laws to explain why the surfboard continues forward after the car brakes
suddenly.
2000 Question 5
The path of the surfboard as it continues over the cliff may be modelled as projectile motion without air
resistance. Calculate the time from the instant the surfboard breaks free until it hits the water.
2000 Question 6
Calculate the horizontal distance travelled by the surfboard, from where it breaks free to where it first strikes
the water.
1999
A car takes off from a ramp and the path of its centre of mass through the air is shown below.
First, model the motion of the car assuming that air resistance is small enough to neglect.
1999 Question 8
Which one of the directions (A–H) best shows the direction of the velocity of the car at point X?
1999 Question 9
Which one of the directions (A–H) best shows the direction of the velocity of the car at point Y?
1999 Question 10
Which one of the directions (A–H) best shows the direction of the acceleration of the car at point X?
1999 Question 11
Which one of the directions (A–H) best shows the direction of the acceleration of the car at point Y?
Now, suppose that air resistance cannot be neglected.
1999 Question 12
Which one of the directions (A–H) could be the direction of the acceleration of the car at point X?
1998
A skateboarder rides up a ramp as shown below. At the instant the skateboard leaves the ramp the centre of
mass of the skateboard and rider is 1.20 m above the ground and is initially moving with a speed of 5.0 ms-1
at an angle of 30° above the horizontal.
The parabola XYZ is the path of the centre of mass of skateboard and rider.
In the following calculations assume that air resistance is negligible.
When the skateboarder is at the highest point of the motion (Y), the speed of the centre of mass is 4.33 ms-1.
1998 Question 11
Calculate the height above the ground of the centre of mass at the highest point of the motion (Y).
When the skateboard touches the ground the centre of mass has moved a horizontal distance of 2.87 m
from point X.
1998 Question 12
Calculate the total time for the centre of mass to travel from X to Z. (Give your answer to three significant
figures.)
1997
In the film Speed a bus travelling at 22 ms-1 is driven over a 15 m gap in an incomplete freeway as shown.
The take-off angle is 11° above the horizontal. In calculating the answers to questions 11 and 12 assume
that air resistance is negligible.
1997 Question 11
After take off, what time does it take for the bus to reach the highest point in its flight?
1997 Question 12
Does the bus land on the other section of the freeway? Show your calculations and reasoning.
Solutions
2000 Solution Q4
Newton’s first law will explain the motion, and that is a body in motion will continue in motion unless a net
force acts on it. The surfboards were travelling at 10ms-1 and were not strapped down therefore they
continued in motion, as there was no net force in the horizontal plane acting on it.
2000 Solution Q5
Since the time taken in this example is determined by the vertical plane, write down all the values you know
for the vertical plane.
x = 80m
u = 0 ms-1
v=?
a = 9.8ms-2
t=?
Then you find one of the equations that involves the three known values and will give you the unknown
value. In this case:
1
x = ut + at 2
2
1
80 = 0 × t + 9.8 × t 2
2
2 × 80
\t=
9.8
t = 4.04s
2000 Solution Q6
Since you now have the time take to hit the ground, and you know that it was travelling at 10ms -1, you simply
Δx
use:
v=
Δt
Manipulate to get:
Δx = v × Δt
and sub the values in.
Δx = 10 × 4.04
Δx = 40.4m
1999 Solution Q8
C
1999 Solution Q9
D
1999 Solution Q10
E
The only force acting is the weight force, so the acceleration is down.
1999 Solution Q11
E
Same as previous question
1999 Solution Q12
F
Now there are two forces acting, the weight force down and the air resistance that is opposing the motion.
1998 Solution Q11
1.52 m
In the vertical direction the initial velocity is vsin300 = 5 × 0.5 = 2.5 m/s.
Use v2 = u2 - 2gh, to find the height.
Since v = 0, (in the vertical direction)
0 = 2.52 - 2 × 9.8 × h
 h = 0.3189 m
You must add this to the height that it started from.
 h = 1.2 + 0.32 = 1.52m
1998 Solution Q12
0.663 secs.
d
d 2.87
t= 
= 0.663 secs.
t
v 4.33
Note that the question specifies the number of sig.fig. that you must include in your answer. The zero at the
start of the answer is not significant.
Since at the top of the flight, the speed is 4.33 m/s. Then v =
1997 Solution Q11
0.43 secs
Note that the examiners use g = 10N/kg.
Consider the vertical component of the velocity.
v = u - gt
where v = 0 and u = 22 sin110
 0 = 22sin110 - 9.8 × t
 t = 4.198  9.8
 t = 0.428
1997 Solution Q12
Yes
You need to assume that the other part of the bridge is the same height as the first side.
If it takes 0.428 secs to get to the top, then it will take 0.8566 secs to be back to the original height.
The horizontal distance travelled in this time = vcos11 × 0.8566
= 22cos 11 × 0.8566
= 18.50 m
DAVID Smith already holds a world record for the greatest distance travelled by a human fired from a
cannon.
But he added to his list of cannonball achievements yesterday by shooting across the US-Mexico border.
And it was all in the name of art. The feat was the brainchild of a Venezuelan artist, Javier Tellez, and is part
of a series of public art projects. Smith climbed into the barrel of the cannon and flashed his US passport as
about 600 people applauded.
He took flight from a popular beach in Tijuana, Mexico, and soared about 45 metres over a line of black
metal poles. He landed uninjured in a net in Border Field State Park in San Diego, with US Border Patrol
agents and an ambulance waiting.
His son, David Smith jnr, is also an accomplished human cannonball. He said his father's flight was the first
across a border by cannon.
Tellez organised the cannonball launch with psychiatric patients at the Baja California Mental Health Centre
in Mexicali, Mexico, as a therapeutic project.
The event is part of an art series that started yesterday, and will run for three months, sponsored by inSite05,
a two-nation arts partnership in the San Diego-Tijuana region.
Tellez called the project "living sculpture", and said it was about "dissolving borders" between the US and
Mexico, and between mental health patients and the rest of the world. "David Smith is a metaphor for flying
over human borders, flying over the law, flying over everything that is established," he said.
Tellez, 36, and Smith snr worked closely on the backdrop, music, costumes and advertising for the project,
One Flew Over the Void. Tellez plans a documentary film about it.
Although it is against the law for anyone, including US citizens, to enter the country outside an official port of
entry, Smith was not crossing illegally. Authorities made an exception for him. Smith snr is listed in Guinness
World Records for the record distance for a human fired from a cannon.
The Smith family has five cannonballs: father, son, two daughters and a cousin. Smith snr built seven
cannons designed to fire humans.
The Age 29.08.05