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• Under special circumstances, we can
use a series of three equations to
describe or predict movement
Vf = Vi + at
d = Vit + 1/2at2
Vf
2 = Vi
2 + 2ad
– Most often, these equations are used to
describe either horizontal or vertical motion
– Acceleration must be constant
Uniformly Accelerated Motion
• Projectile Motion: motion of airborne
stuff
– Horizontal acceleration is zero
• No horizontal forces (neglecting air resistance)
• Therefore, horizontal velocity does not change
– Vertical acceleration is equal to the acceleration
due to gravity (g = –9.8 m/s2)
• Constant attractive force (Newton’s Law of Gravitation)
• Therefore, vertical velocity changes (specifically, it
decreases) at a constant rate (–9.8 m/s every sec)
Projectile Motion: a special case of
uniformly accelerated motion
Projectile Motion
airborne motion -- parabolic path of CM
constant horizontal velocity
constant vertical acceleration
Vertical
velocity = 0
• A ball is dropped from the top of a
building. Its motion may be predicted
using the previous equations.
Time
(s)
Position
(m)
Velocity
(m/s)
Acceleration
(m/s2
)
1 d = Vf = –9.8
2 d = Vf = –19.6
3 d = –44.1 Vf =
4 d = Vf =
Example
Important Characteristics of
Projectile Motion
• Center of mass (CM) of projectile will travel in a
parabolic path - regardless of the motion of the
individual body segments.
• Vertical velocity at the peak of the projectile’s flight will
be exactly zero.
• Horizontal velocity is constant (ignoring air resistance)
• Differences between one- and two-foot
vertical jumping performances
– One-foot flight height: 0.45 m
– Two-foot flight height: 0.54 m
• What were the vertical velocities of the
athlete’s CM at takeoff for each jump?
– What information do we have?
– What information do we want?
– What equation(s) will help us get there?
Example: Vertical Jumping
One-foot jumps
• know:
• want:
• use:
• solution:
Factors Which Influence
Projectile Motion
• Velocity of release
– Magnitude of the velocity (= speed): how fast?
– Angle of release: at what orientation?
• Relative height of release
– from what height was the projectile released?
– at what height did the projectile land?
• Combined, these three factors (speed,
angle, and height) determine how fast,
how high, how long, and how far a
j til ill t l
‘relative
height’
‘speed of
release’
‘angle of
release’
velocity is a vector
it has a horizontal
and vertical component
Two Types of Problems
• Height of takeoff =
height of landing
– Example: kicking a
soccer ball on a level
field for maximum
distance.
• timeup = timedown
• vertical speed at
takeoff = vertical
speed at landing
• Height of takeoff ≠
height of landing
– Example: shooting a
basketball.
• timeup > timedown
• vertical speed at
takeoff > vertical
speed at basket
Importance of Speed, Angle,
and Height of Release
• Speed of release: most important
– Increases in VH increase distance.
– Increases in VV increase time of flight.
• Height of release
– Increases time of flight. VH and VV remain same
• Angle of release
– Affects ratio of horizontal and vertical velocities.
Overall, effect is minimal since increases in one are
offset by decreases in the other.
Factors affecting the ‘Range”
of a projectile
SPEED OF
RELEASE
Speed of Release
• Increases in speed of
release increase two
release parameters in
the same direction:
dH = VH · tTotal
tTotal = 2 ·VV/g
Speed of Release
VHVH
VV
VV
VR
VR
HEIGHT OF RELEASE
Height of Release
• In most circumstances,
increases in height of
release will only affect
time down.
• Therefore, only a portion
of one parameter is
increased
tTotal = tup + tdown
Height of Release
VH
VV
VR
VH
VV
VR
ANGLE OF RELEASE
This graph is WRONG!!! Do you know why?
VH
VV
VR
VH
VV
VR
VH
VV
VR
Steep angle Flat angle
Original angle
Optimum angle of release
• If the height of takeoff is equal to the
height of landing, the optimum angle of
release is always 45 degrees.
• If the height of takeoff is greater than
the height of landing
– The optimum angle is always less than 45 deg.
– For any given height of release, increasing the
speed of release results in an optimum angle
which approaches 45 degrees.
– For any given speed of release, increasing the
height of release results in a decrease in the
optimum angle.
For any given height of release, increasing
the speed of release results in an optimum
angle which approaches 45 degrees.
• Increasing the speed of release will increase VH and time of flight (by
increasing VV).
• As speed increases, optimal angle approaches 45 degrees to balance
the positive effects provided by increasing VH and and by increasing
the time of flight (by increasing VV).
• As long as the height of release is positive (> 0), the optimum angle will
be smaller than 45 degrees. It never actually reaches 45 degrees
because there is some increased time of flight from the positive takeoff
height.
For any given speed of release, increasing
the height of release results in a decrease in
the optimum angle.
• Increasing the height of release
will increase time of flight by
increasing tDOWN (the time taken
for the object to travel from the
peak height to landing).
• Therefore, more of the velocity
component should be directed
horizontally.
• Greater vertical velocity is not
necessary because increasing
the time of flight can be
increased by increasing the
height of release. As a result,
the velocity vector can be
oriented more horizontally. That
is, the optimum angle should
get smaller (flatter).
decreasing
optimum
angle
IncreasingHeight
Long Jump
• What is the optimum angle of takeoff for
long jumpers?
Long Jump
• relative height > 0
– optimum angle will be less than 45 degrees
– when consider the speed of takeoff the
optimal angle should be about 42-43
degrees
– however, in reality the angle of takeoff is
about 15-25 degrees
• WHY?
Long Jump
• This is the “Range” Equation
• Notice that R (range) is proportional to v2
• Therefore, velocity most important
component
• Jumpers can achieve 43 degrees at
takeoff but have to slow down
g
ghvvv
R
2)sin(coscossin 22
++
=
θθθθ
Long Jump
• Moving at 10 m/s
– foot not on ground long enough to
generate a large takeoff angle
– so maintain speed and live with a low
takeoff angle
• v is the most important factor in
projectile motion
VALUES FOR HYPOTHETICAL JUMPS
UNDER DIFFERENT CONDITIONS
Speed of Angle of Relative Height
Values for Takeoff Takeoff of Takeoff
Actual Jump Increased 5% Increased 5% Increased 5%
Variable (1) (2) (3) (4)
Speed of
Takeoff 8.90 m/s 9.35 m/s 8.90 m/s 8.90 m/s
Angle of 20 20 21 20
Takeoff
Relative 0.45 m 0.45 m 0.45 m 0.47 m
Ht of
Takeoff
Horizontal 6.23 m 6.77 m 6.39 m 6.27 m
Range
Change in -- 0.54 m 0.16 m 0.04 m
Horiz
Range
Distance 7.00 m 7.54 m 7.16 m 7.04 m
of Jump

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Projectile motion

  • 1. • Under special circumstances, we can use a series of three equations to describe or predict movement Vf = Vi + at d = Vit + 1/2at2 Vf 2 = Vi 2 + 2ad – Most often, these equations are used to describe either horizontal or vertical motion – Acceleration must be constant Uniformly Accelerated Motion • Projectile Motion: motion of airborne stuff – Horizontal acceleration is zero • No horizontal forces (neglecting air resistance) • Therefore, horizontal velocity does not change – Vertical acceleration is equal to the acceleration due to gravity (g = –9.8 m/s2) • Constant attractive force (Newton’s Law of Gravitation) • Therefore, vertical velocity changes (specifically, it decreases) at a constant rate (–9.8 m/s every sec) Projectile Motion: a special case of uniformly accelerated motion
  • 2. Projectile Motion airborne motion -- parabolic path of CM constant horizontal velocity constant vertical acceleration Vertical velocity = 0
  • 3. • A ball is dropped from the top of a building. Its motion may be predicted using the previous equations. Time (s) Position (m) Velocity (m/s) Acceleration (m/s2 ) 1 d = Vf = –9.8 2 d = Vf = –19.6 3 d = –44.1 Vf = 4 d = Vf = Example Important Characteristics of Projectile Motion • Center of mass (CM) of projectile will travel in a parabolic path - regardless of the motion of the individual body segments. • Vertical velocity at the peak of the projectile’s flight will be exactly zero. • Horizontal velocity is constant (ignoring air resistance)
  • 4. • Differences between one- and two-foot vertical jumping performances – One-foot flight height: 0.45 m – Two-foot flight height: 0.54 m • What were the vertical velocities of the athlete’s CM at takeoff for each jump? – What information do we have? – What information do we want? – What equation(s) will help us get there? Example: Vertical Jumping One-foot jumps • know: • want: • use: • solution:
  • 5. Factors Which Influence Projectile Motion • Velocity of release – Magnitude of the velocity (= speed): how fast? – Angle of release: at what orientation? • Relative height of release – from what height was the projectile released? – at what height did the projectile land? • Combined, these three factors (speed, angle, and height) determine how fast, how high, how long, and how far a j til ill t l ‘relative height’ ‘speed of release’ ‘angle of release’
  • 6. velocity is a vector it has a horizontal and vertical component Two Types of Problems • Height of takeoff = height of landing – Example: kicking a soccer ball on a level field for maximum distance. • timeup = timedown • vertical speed at takeoff = vertical speed at landing • Height of takeoff ≠ height of landing – Example: shooting a basketball. • timeup > timedown • vertical speed at takeoff > vertical speed at basket
  • 7. Importance of Speed, Angle, and Height of Release • Speed of release: most important – Increases in VH increase distance. – Increases in VV increase time of flight. • Height of release – Increases time of flight. VH and VV remain same • Angle of release – Affects ratio of horizontal and vertical velocities. Overall, effect is minimal since increases in one are offset by decreases in the other. Factors affecting the ‘Range” of a projectile SPEED OF RELEASE
  • 8. Speed of Release • Increases in speed of release increase two release parameters in the same direction: dH = VH · tTotal tTotal = 2 ·VV/g Speed of Release VHVH VV VV VR VR
  • 9. HEIGHT OF RELEASE Height of Release • In most circumstances, increases in height of release will only affect time down. • Therefore, only a portion of one parameter is increased tTotal = tup + tdown
  • 10. Height of Release VH VV VR VH VV VR ANGLE OF RELEASE This graph is WRONG!!! Do you know why?
  • 11. VH VV VR VH VV VR VH VV VR Steep angle Flat angle Original angle Optimum angle of release • If the height of takeoff is equal to the height of landing, the optimum angle of release is always 45 degrees. • If the height of takeoff is greater than the height of landing – The optimum angle is always less than 45 deg. – For any given height of release, increasing the speed of release results in an optimum angle which approaches 45 degrees. – For any given speed of release, increasing the height of release results in a decrease in the optimum angle.
  • 12. For any given height of release, increasing the speed of release results in an optimum angle which approaches 45 degrees. • Increasing the speed of release will increase VH and time of flight (by increasing VV). • As speed increases, optimal angle approaches 45 degrees to balance the positive effects provided by increasing VH and and by increasing the time of flight (by increasing VV). • As long as the height of release is positive (> 0), the optimum angle will be smaller than 45 degrees. It never actually reaches 45 degrees because there is some increased time of flight from the positive takeoff height. For any given speed of release, increasing the height of release results in a decrease in the optimum angle. • Increasing the height of release will increase time of flight by increasing tDOWN (the time taken for the object to travel from the peak height to landing). • Therefore, more of the velocity component should be directed horizontally. • Greater vertical velocity is not necessary because increasing the time of flight can be increased by increasing the height of release. As a result, the velocity vector can be oriented more horizontally. That is, the optimum angle should get smaller (flatter). decreasing optimum angle IncreasingHeight
  • 13. Long Jump • What is the optimum angle of takeoff for long jumpers?
  • 14. Long Jump • relative height > 0 – optimum angle will be less than 45 degrees – when consider the speed of takeoff the optimal angle should be about 42-43 degrees – however, in reality the angle of takeoff is about 15-25 degrees • WHY? Long Jump • This is the “Range” Equation • Notice that R (range) is proportional to v2 • Therefore, velocity most important component • Jumpers can achieve 43 degrees at takeoff but have to slow down g ghvvv R 2)sin(coscossin 22 ++ = θθθθ
  • 15. Long Jump • Moving at 10 m/s – foot not on ground long enough to generate a large takeoff angle – so maintain speed and live with a low takeoff angle • v is the most important factor in projectile motion VALUES FOR HYPOTHETICAL JUMPS UNDER DIFFERENT CONDITIONS Speed of Angle of Relative Height Values for Takeoff Takeoff of Takeoff Actual Jump Increased 5% Increased 5% Increased 5% Variable (1) (2) (3) (4) Speed of Takeoff 8.90 m/s 9.35 m/s 8.90 m/s 8.90 m/s Angle of 20 20 21 20 Takeoff Relative 0.45 m 0.45 m 0.45 m 0.47 m Ht of Takeoff Horizontal 6.23 m 6.77 m 6.39 m 6.27 m Range Change in -- 0.54 m 0.16 m 0.04 m Horiz Range Distance 7.00 m 7.54 m 7.16 m 7.04 m of Jump