Galileo’s equations of constant acceleration:
Projectile Motion
v2 = v1 + at
d = v1t + ½at2
v22 = v12 + 2ad
With no air resistance, the path followed by a
projectile will be parabolic
d = displacement
v = velocity
a = acceleration
t = time
subscript 1 = initial time
subscript 2 = final time
You may also use the equations below (derivations of
the aforementioned equations). In many instances,
they can save you time, but are not absolutely
necessary.
Vertical Velocity (Vy)
Horizontal Velocity (Vx)
Horizontal and vertical components of projectile
motion should be treated independently
Gravity will cause the vertical component of
velocity to decrease during the flight at a rate of
9.81 m/s/s
However, horizontal velocity is constant
throughout flight. Why?
Because we are choosing to ignore the effects of
air resistance.
vertical
velocity = 0
Vertical Motion
Vertical Motion
Vertical motion of a projectile (e.g., a water balloon
that is dropped from the edge of a building) is very
predictable
dy = v1yt + ½ayt2
v2y = v1y + ayt
dy = v1yt + ½ayt2
v2y = v1y + ayt
0m
0m
-10 m
-10 m
-20 m
-20 m
time position velocity acceleration
-30 m
time position velocity acceleration
0s
-40 m
0s
1s
-50 m
1s
2s
-60 m
3s
-30 m
0 m/s
−9.8 m/s2
-40 m
−9.8 m/s
m/s2
-50 m
2 s −19.6 m −19.6 m/s
−9.8 m/s2
-60 m
3 s −44.1 m −29.4 m/s
−9.8 m/s2
0m
−4.9 m
−9.8
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Another example
Another example
How high was the cliff? What was dy?
How high was the cliff (what was dy)?
t = ~ 2 s, and using
dy = V1yt + ½ayt2, we
learned that:
dy = ½ · -9.81m/s2 · 2s2
dy = -19.6 m
Perhaps, more importantly, what was V2y?
V2y = -9.81m/s2 · 2s = -19.62 m/s, or -44 mph
Horizontal Motion
Horizontal Displacement
The horizontal displacement (dx), or
range of a projectile, is the main
performance index in numerous
athletic contexts.
dX = VX × t, allows us to simply consider what
factors indluence dX, for a projectile (VX × t). If you
want to consider what influences dX, first consider
VX. What factors affect VX?
Remember, if air resistance is
negligible, there is no net force in
the horizontal direction
(ΣFx = 0; ax = 0)
Given the equation: dx = v1xt + ½axt2,
we can assume that: dX = VX × t
Horizontal Displacement
Now, what affects time dX = VX × t?
1. Vertical speed at release: affected by direction and
magnitude of force that is associated with release
Direction and
magnitude of force
that is associated
with release
Some Application
With all of this in mind, there are
three factors that can be
manipulated to influence dx for a
projectile.
2. Height of release
(if takeoff height = landing height, then tUP = tDOWN)
Relative Height of Release
Speed of Release
Angle of Release
tUP
tDOWN
Which of these factors is most
important?
2
What is most important?
What is most important?
Does increasing height of release always lead to
greater dx?
dX is most sensitive to speed of release…
dx = vx × t
Yes, why? dx = vx × t
Does increasing speed of release always lead to
greater dx?
Yes, but again, why? dx = vx × t
Does increasing angle of release always
lead to greater dx?
It depends, why? dx = vx × t
Of height, speed, and angle of release, speed is the
only factor that can simultaneously and positively
affect vX and t
What are the performance-related
implications of this information?
Theory and Practice
Theory and Practice
How do actual release angles compare to
theoretical optima?
Long Jump:
Positive height of release, so optimal angle should be
slightly lower than 45°
Shot Put:
• Positive height of release, so optimal angle
should be slightly lower than 45°
• Theoretically optimal angle is about 40-41°
• Skilled shot-putters use angles of 31-36°
• Close, but why the discrepancy?
Theory and Practice
Some Practice: The Shot Put
How far will the shot travel?
Long Jump:
Remember: dx = vx × t
• Theoretically optimal angle is about 42°
• Top long jumpers use angles of 17-23°
• Very different. Why the major discrepancy?
Case I
When traveling at ~10 m/s, there is not enough time to
generate a large takeoff angle
Long jumpers sacrifice optimal angle to maximize
horizontal velocity.
speed of release = 12 m/s
height of release = 0 m
angle of release = 30°°
12 m/s
0m
30°°
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Some Practice: The Shot Put
Some Practice: The Shot Put
Use dx = vxt
Find appropriate equation for constant acceleration
need to know vx and t
vx = v cos θ = 12 m/s ⋅ cos 30°° = 10.4 m/s
What is total time in the air (tTOT)?
If height of release = 0, then tUP = tDOWN
and tTOT = tUP + tDOWN
What else do we know?
For upward part of flight:
v2y = v1y + ayt
Plug in v1y, v2y, and ay, then solve for t:
0 m/s = 6 m/s + (−9.81m/s2)(tUP)
tUP = 0.61 s
tTOT will be 2 × tUP: tTOT = 2 × 0.61 s = 1.22 s
(Why ?)
v1 = vy = v · sin θ = 12 m/s ⋅ sin 30°° = 6 m/s
ay = −9.81 m/s2
v2y = 0 m/s
dx = vx tTOT = 10.4 m/s × 1.22 s = 12.7 m
Some Practice: The Shot Put
Some Practice: The Shot Put
Lets use a more realistic height of release
So, the shot traveled 12.7 m horizontally
We still use dx = vxt , but now tUP ≠ tDOWN
vx is found the same way as before:
vx = v cos θ = 12 m/s ⋅ cos 30°° = 10.4 m/s
Case II
speed of release = 12 m/s
height of release = 2.1 m
angle of release = 30°°
12 m/s
30°°
2.1 m
and tUP is also the same (using v2 = v1 + at):
0 m/s = 6 m/s + (−9.81m/s2)(tUP)
tUP = 0.61 s
So how do we calculate tDOWN?
We need to find the upward (dUP) and
downward (dDOWN) displacements
Some Practice: The Shot Put
Some Practice: The Shot Put
Find another appropriate equation
Now calcuate tTOT
d = v1t + ½at2
Calculate the upwards displacement:
dUP = (6 m/s)(0.61 s) + ½ (−9.81 m/s2)(0.61 s)2 = 1.83 m
dDOWN = dUP + ht of release = 1.83 m + 2.1 m = 3.93 m
tTOT = tUP + tDOWN = 0.61 s + 0.90 s = 1.51 s
and finally, calculate the horizontal displacement:
dx = vx tTOT = 10.4 m/s × 1.51 s = 15.7 m
So, now the shot travels 15.7 m horizontally
but this is in the negative direction, so it is −3.93 m
Now find tDOWN:
−3.93 m = (0 m/s) tDOWN + ½ (−9.81 m/s2)(tDOWN)2
0.8012 s2 = (tDOWN)2 , so tDOWN = 0.90 s
The increased height of release resulted in a
3.0 m (~20%) improvement in performance!
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