Projectile Motion
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Transcript of Projectile Motion
Projectile Motion
Projectile motion: a combination of horizontal motion with constant horizontal velocity and vertical motion with a constant downward acceleration due to gravity. Projectile motion refers to the motion of an object that is thrown, or projected, into the air at an angle. We restrict ourselves to objects thrown near the Earth’s surface as the distance traveled and the maximum height above the Earth are small compared to the Earth’s radius so that gravity can be considered to be constant.
Horizontal Component of Velocity
Horizontal and Vertical
Projectile Motion
The motion of a projectile is determined only by the object’s initial velocity and gravity. The vertical motion of a projected object is independent of its horizontal motion. The vertical motion of a projectile is nothing more than free fall.The one common variable between the horizontal and vertical motions is time.
Path of a Projectile
A projectile moves horizontally with constant velocity while being accelerated vertically. A right angle exists between the direction of the horizontal and vertical motion; the resultant motion in these two dimensions is a curved path. The path of a projectile is called its trajectory.
The trajectory of a projectile in free fall is a parabola.
Path of a Projectile
Path of a Projectile
vo = initial velocity or resultant velocity
vx = horizontal velocityvyi = initial vertical velocityvyf = final vertical velocityR= maximum horizontal distance (range)x = horizontal distancey = change in vertical positionyi = initial vertical positionyf = final vertical position = angle of projection (launch angle)H = maximum heightg = gravity = 9.8 m/s2
Path of a Projectile
Path of a Projectile
The horizontal distance traveled by a projectile is determined by the horizontal velocity and the time the projectile remains in the air. The time the projectile remains in the air is dependent upon gravity. Immediately after release of the projectile, the force of gravity begins to accelerate the projectile vertically towards the Earth’s center of gravity.
Path of a Projectile
The velocity vector vo changes with time in both magnitude and direction. This change is the result of acceleration in the negative y direction (due to gravity). The horizontal component (x component) of the velocity vo remains constant over time because there is no acceleration along the horizontal direction
The vertical component (vy) of the velocity vo is zero at the peak of the trajectory. However, there is a horizontal component of velocity, vx,
at the peak of the trajectory.
Path of a Projectile
Path of a ProjectileIn the prior diagram, r is the position vector of the projectile. The position vector has x and y components and is the hypotenuse of the right triangle formed when the x and y components are plotted. The velocity vector vot would be the displacement
of the projectile if gravity were not acting on the projectile.
The vector 0.5gt2 is the vertical displacement of the projectile due to the downward acceleration of gravity.
Together, this determines the vertical position for the projectile:
Δy = (vy·t) – (0.5·g·t2)
Path of a Projectile
Acceleration of a Projectile
g
g
g
g
g
x
y
Acceleration points down at 9.8 m/s2 for the entire trajectory of all projectiles.
Velocity of a projectile
vo
vf
v
v
v
x
y
Velocity is tangent to the path for the entire trajectory.
Position graphs for 2-D projectiles
x
y
t
y
t
x
Velocity graphs for 2-D projectiles
t
Vy
t
Vx
Acceleration graphs for 2-D projectiles
t
ay
t
ax
Fig. 04.24
Problem Solving: Projectile Motion
Analyze the horizontal motion and the vertical motion separately. If you are given the velocity of projection, vo, you may want to resolve it into its x and y components. Think for a minute before jumping into the equations. Remember that vx remains constant throughout the trajectory, and that vy = 0 m/s at the highest point of any trajectory that returns downward.
Horizontal velocity component:
vx is constant because there is no acceleration in the horizontal direction if air resistance is ignored.
cos ox vv
Vertical velocity component:
At the time of launch:
After the launch:
If vy positive, direction of vertical motion is up; if vy negative, direction of vertical motion is down; if vy = 0, projectile is at highest point.
sin oyi vv
tgvv oyf sin
Horizontal position component:
If you launch the projectile horizontally: then vo = vx
vyi = 0 m/s = 0o
tθcosvx
tvx
o
x
Vertical position component:
2oif
2yiif
if
tg0.5tθsinvyy
tg0.5tvyy
yyΔy
Relationship Between Vertical and Horizontal Position:
this equation is only valid for launch angles in the range 0 < < 90
22o
2
θcosv2
gxθtanxyΔ
Range (total horizontal displacement)
g
θcosθsin2vR
θcosθsin2θ)(2sin
:formulaangledouble
g
θ2sinvR
2o
2o
Maximum Height
g2
θsinvH
22o
When Do The Range & Maximum Height Equations Work?
Works when y = 0.
Does not work when y 0.
Determining vo from vx and vy
If the vertical and horizontal components of the velocity are known, then the magnitude and direction of the resultant velocity can be determined.Magnitude: 22
yxo vvv
Determining vo from vx and vy
Direction: from the horizontal
Direction: from the vertical
x
y1
v
vtanθ
y
x1
v
vtanθ
Range and Angle of Projection
Range and Angle of Projection
The range is a maximum at 45 because sin (2·45) = 1.For any angle other than 45, a point having coordinates (x,0) can be reached by using either one of two complimentary angles for , such as 15 and 75 or 30 and 60 .
Range and Angle of Projection
The maximum height and time of flight differ for the two trajectories having the same coordinates (x, 0).
A launch angle of 90° (straight up) will result in the maximum height any projectile can reach.
For Objects Shot Horizontally:
vx constant
y negative; y = - height
tvxsm
0vtx
v
x
yix
2tg5.0y
Zero Launch angle
A zero launch angle implies a perfectly horizontal launch.
vo
For Objects Shot Horizontally:
When hits at bottom: Vyf should be negative vo = resultant velocity
2yf
2xo
yf
vvv
tgv
For Objects Shot Horizontally:
with horizontal:
with vertical:
x
yf1
v
vtan
yf
x1
vv
tan
For Situations in Which y = 0 m
2yi
x
ox
oyi
22o
2o
tg5.0tvy
tR
v
cosvv
sinvv
g2sinv
H
g2sinv
R
For Situations In Which y Positive
For Situations In Which y Positive
At any point in the flight:
sinvvcosvv oyiox
2yf
2xo
x
x
yiyf
22o
2
2yi
vvv
ttanconsv
tvx
tgvv
cosv2
gxtanxy
tg5.0tvy
For Situations In Which y Negative
For Situations In Which y Negative
At launch:After launch:
When it hits ground:
sinvvcosvv oyiox
22o
2
2yi
x
cosv2
gxtanxy
tg5.0tvy
tvx
2
yf2
xo
yiyf
x
vvv
tgvv
ttanconsv