Projectile Motion
I. 2-Dimensional Motion
• 2-dimensional motion simply means that the motion is caused by a force or forces that have both an x and y component (remember vector addition?)
• The first situations we’ll study will involve two forces, one acting vertically and one horizontally
• Projectile motion is influenced first by the force that projects the object, and then only gravity, which pulls it vertically downward
• Circular motion consists of a force that pulls the object toward the center of motion, and a force that wants the object to move in a straight line.
• A 2-dimensional force is exerted at an angle to the horizontal. It can be separated into two perpendicular components that are independent of each other
ANY FORCE VECTOR EXERTED AT AN ANGLE TO THE HORIZONTAL WILL HAVE A RESOLVABLE X AND Y COMPONENT OF FORCE
THE VECTOR SUM OF THESE COMPONENTS GIVES THE FORCE EXERTED AT THE GIVEN ANGLE
1 Vector Resolution Quick Review Suppose you are pulling a box with a rope… R The force vector can be resolved into an x and y component.
Ry R R Ry
Rx Rx the vectors are related to each other by a right triangle when the x and y components are added head to tail
The force vectors Rx and Ry are components of the force vector R.
This force analysis says that a force applied at an angle gives a vertical force and a horizontal force
A common feature of 2-D force problems will be to resolve the forces, make a right triangle, and analyze the triangle. 2 HOW TO WORK A PROBLEM
1. Read the problem CAREFULLY 2. Get a mental picture of the situation 3. Identify the variables you are solving for 4. DRAW A PICTURE 5. Analyze the picture with vectors, labeled parts, etc 6. Break out any parts of the picture you need such as vectors to add, right triangles, etc 7. Break out any horizontal and vertical components and separate them 8. Decide which equations will help you solve the problem. 9. Often, just looking at the UNITS will help you identify the equations 10. Show all of your work NEATLY 11. The answer must have the correct units or it is WRONG.
NOTICE THAT WE DON’T THINK ABOUT THE EQUATIONS FIRST. WE THINK ABOUT THEM LAST.
3 II. Projectile Motion
A. DEFINITIONS
• By DEFINITION, the MOTION of a PROJECTILE is influenced only by the force of GRAVITY.
• This is like free fall EXCEPT that PROJECTILE MOTION has a HORIZONTAL direction and a VERTICAL direction
• A projectile is NOT a rocket or any other object with its own source of thrust
• The SHAPE of the path in projectile motion is always a PARABOLA (unless the projectile is dropped straight down or thrown straight up)
• Projectile motion is something a lot of people think they can describe, but their understanding has misconceptions. • The most common misconception is that a moving object needs a force to KEEP it in motion. We know this is NOT so (right?)
• Another misconception involves the forces acting on it. After the initial thrust or push to launch a projectile ONLY GRAVITY ACTS ON IT (we’ll ignore air resistance)
• Another misconception is the time it takes to hit the ground. People think that the horizontal motion keeps the projectile in the air longer. If I fire a projectile HORIZONTALLY (no angle up or down), it will hit the ground at the same time as if I had just dropped the projectile from the same initial height 4 B. Characteristics of projectile motion
Projectile motion has horizontal and vertical components
horizontal motion
projectile
vertical motion
WHAT DO YOU THINK I’M TRYING TO TELL YOU WITH THIS DIAGRAM? 5 horizontal projectile motion Look at the vector lengths..all the same.
Velocity is constant, vertical projectile motion Acceleration is zero (aka FREE FALL) Look at the vector lengths… they get bigger. Velocity increases, object accelerates Suppose we add the vertical and horizontal components…place them head to tail…the resultant is the projectile motion of the launched object
6 SAY IT AGAIN… in the absence of gravity, horizontal motion at CONSTANT VELOCITY (zero acceleration) is the only motion path
with gravity, a vertical motion component is added to the horizontal component. BUT, If the projectile is pushed off the edge, the vertical component it only has a vertical motion path due ACCELERATES down due to gravity. It ACCELERATES downward. to gravity.
THE VERTICAL AND HORIZONTAL COMPONENTS ACT INDEPENDENTLY. THE VERTICAL COMPONENT ACCELERATES, THE HORIZONTAL COMPONENT DOES NOT 7 REMEMBER. ONLY GRAVITY IS INFLUENCING A LAUNCHED PROJECTILE.
whether dropped or launched, these objects hit the ground at the same time. If the object is launched UP or DOWN, then this is no longer true.
8 C. Launching at an angle Launching horizontally or at an angle…
Either way
THE HORIZONTAL VELOCITY DOES NOT CHANGE THE VERTICAL VELOCITY CHANGES BY 9.8 m/s EVERY SECOND
images from the physics classroom on-line
horizontal vertical
Look closely at the numbers 9 Can you see that the vertical component of the motion (velocity) is the only one that changes? It changes by -9.8 m/s every second. That is the acceleration due to gravity.
Can you also see that there is no vertical component when the projectile reaches its MAXIMUM HEIGHT? 10 I can tell from the picture that the cannon was fired with an initial velocity of
R 2 2 m θ 19.6 m/s R = 19.6 + 73.1 = 75.7 s 73.1 m/s 19.6 Also, the angle of the cannon must be tanθ = = 0.268 73.1 θ = tan−1 0.268 = 15.0o
11 D. TIME AND MAXIMUM HEIGHT
maximum height
ymax x/2 X t = 0 t = ttot launch hit
ttot = 2ty max
The time at the projectile’s maximum height is half the total flight time
12 E. 2-D KINEMATIC EQUATIONS (for projectiles launched at an angle)
Since the motion has 2-dimensions, the equations break out into a set of horizontal and vertical equations. Otherwise, they are JUST LIKE the 1-D equations from the previous unit.
HORIZONTAL VERTICAL 1 1 x = v t + a t 2 y = v t + gt 2 ix 2 x iy 2
v fx = vix + axt v fy = viy + gt 2 2 2 2 v fx = vix + 2ax x v fy = viy + 2gy
x is horizontal displacement y is vertical displacement g is acceleration due to gravity 2 since there is no horizontal acceleration, ax = 0 m/s
x = vixt
v fx = vix v2 = v2 these two are trivial. They just reinforce the idea that fx ix horizontal velocity is constant 13 Sample Problem…
A football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the horizontal. Determine the time of flight, the horizontal displacement, and the peak height of the football.
25 m/s ymax 45o
x and ttot
Break out some vectors We want to know: time of flight: ttot 25 m/s vy horizontal displacement: x 45o
vx peak height: ymax
We can’t get horizontal distance without vx We can’t get vertical height without vy
14 o the initial direction of vix is + vix = 25cos 45 m 25 m/s the initial direction of viy is + vix = 17.7 viy s o o vfx = vix 45 viy = 25sin 45 v m v = -v this is because of the shape ix v = 17.7 fy iy iy s of a parabola. The direction changes after the halfway point in the flight path
The horizontal distance only depends on vx and time The vertical height depends on vy, g, and time
Time is missing from both solutions. Let’s try to solve for it first.
v fy = viy + gt v fy = viy + gt Solve for t m m m −17.7 = 17.7 + (−9.8 )t s s s2 Time of flight t = 3.6s
15 It’s EASY from here…
ttot = 3.6 s
x = vixt m x = (17.7 )3.6s horizontal displacement: s x = 63.7m
peak height: t = 2t 1 tot y max y = v t + gt 2 iy 2 ttot t = m 1 m y max 2 y = (17.7 )1.8s + (−9.8 )1.82 s 2 s2 3.6s t = y = 16.0m y max 2
ty max = 1.8s
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