Projectile Motion. A Projectile What is a projectile? –A projectile is any object that is acted...

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Transcript of Projectile Motion. A Projectile What is a projectile? –A projectile is any object that is acted...
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 y
direction.– 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 yaxis for vertical motion where gravity is
a factor (ay = g = 9.81 m/s2).
– Choose the xaxis for horizontal motion. Since there are no forces acting in this direction, the speed will be constant (ax = 0).
– Analyze motion along the yaxis separate, or independently from motion along the xaxis. 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 ydirection, you also have it for the xdirection, and viceversa.
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.
iyixi
ix
ix
ix
iy
ix
vf = vi
Formulas You Will Use
• xdirection:– ddxx = v = vix ix t = vt = vfxfxt t (velocity is constant since a(velocity is constant since axx = 0) = 0)
– vvixix = v = viicoscos
• ydirection:ydirection:– ddyy = ½ (v = ½ (vii + v + vff) t = v) t = vavgavg t t
– vvfyfy = v = viyiy + gt + gt
– ddyy = v = viyiy t + ½ g(t) t + ½ g(t)22
– vfy2 = viy
2 + 2gdd
– vviyiy = v = viisinsin
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 ydirection while it is constant in
the xdirection.
– Acceleration in the y direction only.
vy (Increasing)
vx (constant)
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.
35
v i = 8.5 m
/s
viy
vix
Example #1 (Part 1  cont.):• Using trigonometry (see reference table).
– viy = visin = (8.5 m/s)(sin 35) = 4.9 m/s
– vix = vicos = (8.5 m/s)(cos 35) = 7.0 m/s
35
v i = 8.5 m
/s
viy
vix
Example #1 (Part 2): Finding the Maximum Height• Whenever a projectile motion question asks about anything related to motion in the vertical direction, you should automatically know that you will need a 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
• 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: vfy
2 = viy2 + 2gdd
(0 m/s)2 = (4.9 m/s)2 – (2)(9.81 m/s2)(d)
• Solving for d yields:d = 1.2 m
viy vfy ay = g dy t4.9 m/s 0 m/s 9.81 m/s2 ? ?
(1) vf = vi + at(2) d = vit + ½ at2
(3) v2 = vi2 +2ad
Example #1 (Part 3): Finding the time of flight• As before, list all the known and
unknown variables.
• 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
(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 ?
Example #1 (Part 3): An Alternative Way• Instead of using equation (1)
we will use equation (2) this time.• 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:
• 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.
(1) vf = vi + at(2) d = vit + ½ at2
(3) v2 = vi2 +2ad
ssm
m
g
dt y 49.0
/81.9
2.1222
Example #1 (Part 4): Finding the Horizontal Distance
• To find the horizontal distance, you only need to be concerned with motion in the xdirection.
• As previously mentioned, there is no acceleration in the xdirection, therefore, vix = vfx = vx = constant.
• Equation (2) can be used since the portion containing the acceleration (½ axt2) will be zero. Hence:
dx = vixt
• Since the initial velocity in the xdirection and the time have already been determined:
dx = (7.0 m/s)(1.0 s) = 7.0 m
(1) vf = vi + at(2) d = vit + ½ at2
(3) v2 = vi2 +2ad
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.
• Since we want to find dy, we will have to use either equation (2) or (3).Equation (2) is a better fit since wedo 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
viy vfy ay = g dy t
0 m/s ? 9.81 m/s2 ? 2.5 s
(1) vf = vi + at(2) d = vit + ½ at2
(3) v2 = vi2 +2ad
Example #2(Part 4): Determining the final velocity.
• Start by listing all variables known.
• 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
vfy = (0 m/s)(2.5 s) – (9.81m/s2)(2.5 s)
vfy = 24.5 m/s
vix = vfx viy vfy ay = g dy t
30 m/s 0 m/s ? 9.81 m/s2 31 m 2.5 s
(1) vf = vi + at(2) d = vit + ½ at2
(3) v2 = vi2 +2ad
Example #2(Part 4): Determining the final velocity.
• 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.
• Applying the Pythagorean Theorem, we get:
vf2 = vfx
2 + vfy2
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
smv
smsmv
vvv
f
f
fyfxf
/7.38
)/5.24()/30( 22
22
vf = ?
vfx
vfy