Week 9 Flashcards
PROJECTILE MOTION
• A PROJECTILE is a body in free fall that is only subjected to forces of GRAVITY and AIR RESISTANCE (An aeroplane is not a projectile!)
• Vertical (gravity and height) and Horizontal (displacement - range) components <>
• Independent of one another
• Sporting Relevance - manipulate, control or assess flight
path
• Football, Tennis etc
• Horizontal Displacement - Long Jump, shot put, javelin, discus, hammer
• Vertical Displacement - high jump, pole vault
TERMINOLOGY
The flight path of a projectile is its Trajectory (gravity continually acts to change motion)
The shape of the flight path is called a Parabola
The highest point in the trajectory is called the Apex and the horizontal displacement is the Range
3 FORCES INFLUENCING PROJECTILE MOTION
Applied force(verticle/horizontal)
Initial applied force = resultant force
Air resistance
Gravity
VERTICAL & HORIZONTAL COMPONENTS
Once a body has been projected into the air, its overall (resultant) velocity is constantly changing but the vertical and horizontal components change in a predictable manner
• Horizontal Velocity Component
• Air resistance – usually ignored
• Horizontal Velocity is constant during flight
• Vertical Velocity Component
• upward velocity = positive, decreases to apex
• downward velocity = negative, increases from apex • Vertical velocity at apex = 0
• Acceleration
• Horizontal = 0 (if air resistance is negligible) • Vertical(g) =-9.81m·s-2
VERTICAL KINEMATICS OF CENTRE OF MASS DURING FLIGHT OF STANDING VERTICAL JUMP
Position
Increases to apex Decreases from apex No horizontal motion
Velocity
Constant rate of change
Positive velocity decreases to zero at
apex
Negative velocity increases from zero at apex
Acceleration
Constant at -9.81 ms-2
Rate of change in velocity
HORIZONTAL VELOCITY DURING STANDING LONG JUMP
Horizontal velocity remains constant whilst airborne
VERTICAL VELOCITY DURING STANDING LONG JUMP
Vertical velocity changes whilst airborne, because of the
constant downward acceleration due to gravity
- ANGLE OF RELEASE
The direction a body is projected relative to the horizontal”
• Changing angle for a constant release velocity affects maximum Height and Range
• 3 scenarios
• Vertical trajectory 90° = vertical motion, 0 horizontal velocity
• Oblique Trajectory - any angle between 0 -90°
• Horizontal Trajectory 0° = horizontal motion, 0 vertical velocity.
Optimum angle for Distance = 45°
• High Jump = 40 - 48°
• Long Jump = 18 - 27°
• Basketball <4.57m from basket = 52 - 55°
• Basketball at 6.40m = 48 - 50°
• Ski Jump = 4.6 - 6.2°
Long jump forward velocity > upwards velocity
PROJECTION VELOCITY
For a constant release angle, changing the velocity affects maximum height and range. – Vertical component affected by acceleration due to gravity (-9.81ms-2)
– Horizontal component constant throughout flight
– The relationship between VELOCITY and RANGE is curvilinear, with Range proportional
to Velocity2
RELATIVE PROJECTION HEIGHT
Affects RANGE and FLIGHT TIME
Height of release greater than height of landing - Optimal angle <45° e.g. shot putt (38-42°)
Height of release lower than height of landing - Optimal angle >45° e.g. maximum distance basketball shot >5 m away (48-50°)
Based in the positioning of the body , joint positioning at the point of take of and landing, calculations could have a little bit of error in them
ANALYSING PROJECTILE MOTION
Initial velocity of projectile = initial speed (magnitude) and the angle of projection (direction)
Therefore we can use
Trigonometry and Pythagoras Vertical Velocity = vsin
Horizontal Velocity = vcos
Can also use Equations of Motion as acceleration
is constant at - 9.81ms-2 or zero
When release height and landing occur on the same level time up = time down
The flight time will depend on the vertical velocity at release and the height of release above the landing surface.
EQUATIONS OF MOTION
For motion with constant acceleration the following relationships apply:
V= u+at
S=tu+ 1/2at2
V2=u2+2as
For projectile motion, a is constant at -9.81ms-2
u = initial velocity
v = final velocity
s = displacement
a = acceleration
t = time
CHANGES IN POSITION
• If, at time t=0, we started tracking the movement of the centre of mass and its coordinates were: x(0)=x0 , y(0)=y0 , and z(0)=z0
• We can generate the formula for each of the coordinates as functions of time by finding the areas beneath the corresponding velocity components drawn against time
AERIAL PHASE OF A VERTICAL JUMP
SOLVING PROJECTILE MOTION PROBLEMS
Write down all given values (e.g. Inital velocity)
✓Write down all the rules you know for projectile motion
• Vertical and Horizontal components of velocity from vSinθ and vCosθ respectively • Vertically – v at apex 0, a= -9.81ms-2, tup = tdown
• Horizontal – v is constant (u=v, a=0)
✓To find HEIGHT or TIME • useverticalmotion
✓To find RANGE
➢ use horizontal motion (d=ut)
KEY POINTS
Projectile motion is defined by applied force, gravity and air resistance
• Flight path is determined by release angle, release velocity and release height •Projectile problems can be solved by considering vertical and horizontal components • Vertical acceleration is constant at -9.81ms-2 and horizontal acceleration is zero
• Height, time and range can be calculated using the Equations of Motion
