two-dimensional motion chapter 3. a little vocab projectile = any object that moves through space...
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Two-Dimensional MotionChapter 3
A little vocab
Projectile = any object that moves through space acted on only by gravity
Trajectory = the path followed by a projectile
Range = the horizontal distance a projectile travels
Puzzle
If a ball is dropped straight down at the same time one is launched horizontally, which ball will hit the ground first?
Does this work the same if it is a bullet launched horizontally, at the same time a ball is dropped from the same height?
Without any means to propel an object, once it is moving horizontally, it has constant horizontal motion… x = vt
(no acceleration)
In free fall, vertical motion… y = vit + ½ gt2
We know…
Put these motions together
Motion in Two Dimensions
Using + or – signs is not always sufficient to fully describe motion in more than one dimension Vectors can be used to more fully
describe motion
Still interested in displacement, velocity, and acceleration
Projectile Motion
An object may move in both the x and y directions simultaneously It moves in “two dimensions”
The form of two dimensional motion we will deal with is an important special case called projectile motion
Assumptions of Projectile Motion
Because this is a “perfect physics world” We may ignore air resistance and friction We may ignore the rotation of the earth
With these assumptions, an object in projectile motion will follow a parabolic path
Rules of Projectile Motion The x- and y-directions of motion are
completely independent of each other X-motion and Y-motion happen at the
same time ay = g
Why? The y-direction is free fall
ax = 0 Why? The x-direction is uniform motion There is nothing else pushing in the horizontal
direction
Horizontal Projectile Motion
something thrown horizontally Without gravity, it would continue along a
horizontal path. Because gravity acts “downward”, the path is half
of a parabola.
• Start at maximum height
• Given a push in the X-direction only
Solving Horizontal Projectile Motion Make a chart with 2 sides What do we ALREADY know (without even
having a problem to solve?)Horizontal Vertical
a =
vi =
vf =
t =
x =
a =
vi =
vf =
t =
x =
0
#
# = vi
#
# = range
9.8m/s2 = g
0
# (just before it hits)
#
Max height = y
NOTE: Can only cross over at t!
Solving contunued…
Use kinematics equations to solve Horizontal… if a = 0, the only
equation that matters is #2 x = vit
Vertical …1. vf = vi + gt
2. y = vit + ½gt2
3. vf2 = vi
2 + 2gy
HINT: Solve for time first… goes on both sides of chart!
Example
A dart player throws a dart horizontally at a speed of 12.4 m/s. The dart hits the board 32 cm below the height from which it was thrown. How far away is the player from the board?
Horizontal Vertical
a =
vi =
vf =
t =
x =
a =
vi =
vf =
t =
y =
0
12.4 m/s
12.4 m/s
?
?
9.8m/s2
0
?
?
32cm = .32m
What can we solve first? We are looking for “x” Not enough info on the horizontal
side… On the vertical side, we can solve for
vf
t Solve for t, so we can use it on the
horizontal side Don’t really care about vf this time
The math…
For t
y = vit + ½gt2
.32 = 0 + ½ (9.8)t2
t = 0.2556s
Horizontal Vertical
a =
vi =
vf =
t =
x =
a =
vi =
vf =
t =
y =
0
12.4 m/s
12.4 m/s
0.2556s
?
9.8m/s2
0
?
0.2556s
32cm = .32m
For x
x = vt
x = (12.4)(.2556)
x = 3.17m
What if… the previous question asked for the
total final velocity? This is where VECTORS come in… We knew that vf was the same as the
vi in the horizontal = 12.4m/s We can find vf in the vertical
vf = vi + gtvf = 0 + (9.8)(0.2556)vf = 2.5m/s
We aren’t done yet.
Total velocity is a vector in polar coordinates.
12.4m/s
2.5
m/sResultant
(r,Θ)
r2 = 12.42 + 2.52
r = 12.6m/s
tanΘ = -2.5/12.4
Θ = tan-1(-0.2016)
Θ = -11.4o
Θ = -11.4o
Θ