Download v iy

Projectile Motion
A Projectile
• What is a projectile?
– A projectile is any object that is acted upon by
gravity alone.
– Note that gravity acts in the negative ydirection.
– Air resistance is ignored in projectile motion
unless explicitly stated.
– The path of a projectile is parabolic in nature.
The Trajectory of a Projectile
Choosing Coordinates & Strategy
• For projectile motion:
– Choose the y-axis for vertical motion where gravity is
a factor (ay = g = -9.81 m/s2).
– Choose the x-axis for horizontal motion. Since there
are no forces acting in this direction, the speed will be
constant (ax = 0).
– Analyze motion along the y-axis separate, or
independently from motion along the x-axis. This
is a step that most students have difficulty with.
– Note that time is the only variable that will always be
the same for both the vertical and horizontal
directions. Hence, if you can find if for the y-direction,
you also have it for the x-direction, and vice-versa.
Strategies Continued
• If the projectile is fired
horizontally, then viy will be
zero.
• If the projectile is launched at
an angle greater than 0o, then
vy will be 0 m/s at the very
peak of its trajectory.
• A common misconception is
that the acceleration = 0 m/s2
at the peak as well. But it
actually equals -9.81 m/s2, and
anywhere else along the path
of the projectile
• The magnitude of the initial
launch velocity will be the
same as the final velocity if the
projectile lands at the same
height from which it is
launched.
ix
ix
i
ix
iy
ix
ix
iy
vf = v i
Formulas You Will Use
• x-direction:
– dx = vix t = vfxt (velocity is constant since ax = 0)
– vix = vicos
• y-direction:
–
–
–
–
–
dy = ½ (vi + vf) t = vavg t
vfy = viy + gt
dy = viy t + ½ g(t)2
vfy2 = viy2 + 2gd
viy = visin
The Vectors of Projectile Motion
• What vectors exist in projectile motion?
– Velocity in both the x and y directions.
• Note that it increases in the y-direction while it is constant in
the x-direction.
– Acceleration in the y direction only.
vx (constant)
vy (Increasing)
ay = 9.81 m/s2
ax = 0
Example #1: Determining the vertical and horizontal components,
and maximum height of a projectile
A child kicks a soccer ball with an initial velocity of 8.5 meters per second at an
angle of 35º with the horizontal, as shown.
The ball has an initial vertical velocity of 8.5 meters per second at 35 relative
to the horizontal. [Neglect air resistance.]
1. Determine the horizontal and vertical components of the ball’s initial
velocity.
2. Determine the maximum height reached by the ball.
3. Determine the total amount of time the ball is in the air.
4. Determine how far the ball travels before it hits the ground.
Example #1 (Part 1): Determine the
vertical and horizontal components of
the initial velocity.
• For the vertical and horizontal velocities, you need to use
component vector analysis using trigonometry.
• Begin by creating a right triangle with the vertical and
horizontal components making up the a and b legs of the
right triangle.
viy
35
vix
Example #1 (Part 1 - cont.):
• Using trigonometry (see reference table).
– viy = visin = (8.5 m/s)(sin 35) = 4.9 m/s
– vix = vicos = (8.5 m/s)(cos 35) = 7.0 m/s
viy
35
vix
Example #1 (Part 2): Finding the Maximum Height
•
Whenever a projectile motion question
(1) vf = vi + at
asks about anything related to motion in
the vertical direction, you should
(2) d = vit + ½ at2
automatically know that you will need a
(3) v2 = vi2 +2ad
formula involving the acceleration due to
gravity. Substitute g wherever you see a.
• In this example, ask yourself what do you know about the ball’s motion at
the top of its trajectory.
•
•
•
•
– You know that the vertical component of velocity at the peak of the soccer ball’s
trajectory will be 0 m/s. Therefore, we will set vfy = 0 m/s.
In the vertical direction, we know that gravity will act on the soccer ball such
that it will cause it to accelerate at -9.81 m/s2.
List the variables that you know and don’t know
viy
vfy
ay = g
dy
t
4.9 m/s
0 m/s
-9.81 m/s2
?
?
Choose a formula from the three in the table above that contains all the
known variables with only the unknown one missing.
Hence you will use equation (3) from the table above: vfy2 = viy2 + 2gd
(0 m/s)2 = (4.9 m/s)2 – (2)(9.81 m/s2)(d)
Solving for d yields:
d = 1.2 m
Example #1 (Part 3): Finding the time of flight
• As before, list all the known and
unknown variables.
(1) vf = vi + at
(2) d = vit + ½ at2
(3) v2 = vi2 +2ad
viy
vfy
ay = g
dy
t
4.9 m/s
0 m/s
-9.81 m/s2
1.2 m
?
• You have two choices of equations that you can use, (1)
or (2)
• Solving for t using equation (1): vfy = viy + at
(0 m/s) = (4.9 m/s) – (9.81 m/s2)(t)
t = 0.5 s
– However, this time only covers half the flight of the ball,
therefore, the total time is double, or 1.0 s
Example #1 (Part 3): An Alternative Way
• Instead of using equation (1)
(1) vf = vi + at
we will use equation (2) this time. (2) d = vit + ½ at2
2 = v 2 +2ad
(3)
v
i
• We will also analyze the problem
starting from the second half of the trajectory.
Since the velocity is 0 m/s at the top of its
motion, viy = 0 m/s.
• Equation (2) then simplifies to: dy = ½ gt2
• Solving for t gives us: t 
2 1.2m

 0.49s
2
g
9.81m / s
2d y
• As before, this time only accounts for ½ of the
whole trajectory of the ball, therefore you must
double it to get a total time of 1.0 s.
Example #1 (Part 4): Finding the Horizontal
Distance
• To find the horizontal distance, you only need to be
concerned with motion in the x-direction.
• As previously mentioned, there is no acceleration in the
x-direction, therefore, vix = vfx = vx = constant.
• Equation (2) can be used since the portion containing
the acceleration (½ axt2) will be zero. Hence:
(1) vf = vi + at
dx = vixt
(2) d = vit + ½ at2
(3) v2 = vi2 +2ad
• Since the initial velocity in the x-direction and the time
have already been determined:
dx = (7.0 m/s)(1.0 s) = 7.0 m
Example #2: Determining the horizontal distance,
height and final velocity of a projectile launched
horizontally.
• A projectile is launched horizontally at a speed of 30. meters per
second from a platform located a vertical distance h above the
ground. The projectile strikes the ground after time t at horizontal
distance d from the base of the platform. [Neglect friction.]
1. Sketch the theoretical path of the projectile.
2. Calculate the horizontal distance, d, if the projectile’s total time of
flight is 2.5 seconds.
3. Determine the height, h, of the platform.
4. Determine the final velocity, vf, when the projectile hits the ground.
Example #2(Part 1): Sketching the path.
• All projectiles followed a curved or parabolic
path.
Example #2(Part 2): Determining the
Horizontal Distance.
• As in the previous example, the velocity in the
horizontal direction is constant.
dx = vixt
dx = (30 m/s)(2.5 s) = 75 m
Example #2(Part 3): Determining the
height of the platform.
• As in the previous example, we will list those variables
we know and those that we do not.
viy
vfy
ay = g
dy
t
0 m/s
?
-9.81 m/s2
?
2.5 s
• Since we want to find dy, we will
(1) vf = vi + at
have to use either equation (2) or (3).
(2) d = vit + ½ at2
Equation (2) is a better fit since we
(3) v2 = vi2 +2ad
do not know the final velocity yet.
dy = viyt + ½ gt2
dy = (0 m/s)(2.5 s) – ½ (9.81m/s2)(2.5 s)2
dy = 31 m
Example #2(Part 4): Determining the final
velocity.
• Start by listing all variables known.
vix = vfx
viy
vfy
ay = g
dy
t
30 m/s
0 m/s
?
-9.81 m/s2
31 m
2.5 s
• The final velocity, vf, is the sum of both the vertical and
horizontal components of the final velocity, where we will
use the Pythagorean Theorem.
• But first, we need to find the final velocity in the vertical
direction using equation (1)
vfy = viyt + gt
(1) vf = vi + at
vfy = (0 m/s)(2.5 s) – (9.81m/s2)(2.5 s) (2) d = vit + ½ at2
(3) v2 = vi2 +2ad
vfy = -24.5 m/s
Example #2(Part 4): Determining the final
velocity.
vix = vfx
viy
vfy
ay = g
dy
t
30 m/s
0 m/s
-24.5 m/s
-9.81 m/s2
31 m
2.5 s
• A vector diagram can help to fully understand how the final velocity
is the sum of the vector components in the horizontal and vertical
directions.
vfx
• Applying the Pythagorean Theorem, we get:
vf2 = vfx2 + vfy2
v f  v 2fx  v 2fy
v f  (30m / s ) 2  (24.5m / s ) 2
v f  38.7 m / s
vfy
vf = ?