Chapter 3

The Giancoli onLine Tutor

Kinematics in Two Dimensions; Vectors

In this chapter we move from one dimension to two dimensions. The mathematical language for handling multiple dimensions is Vectors. In this chapter we first learn about vectors and then apply them to Projectile Motion.

I. Vectors

Adding Vectors

The best way to add vectors is to use the vector component table. If you want to find the components of a vector R=A+B+C construct the following table:
(A means the vector A and |A| means its magnitude.)
x-componenty-component
A|A|*cosqa|A|*sinqa
B|B|*cosqb|B|*sinqb
C|C|*cosqc|C|*sinqc
RAx+Bx+CxAy+By+Cy

It is important to measure all angles counter-clockwise from east. If the magnitude (length) of A, |A|, in the following figure is 10 and B is 12 then

Ax=10*cos45o= 7.07 and Bx=12*cos135o= -8.48.
When the table is done the magnitude of R is found from using

|R|2=Rx2+Ry2
and its direction by
qR=tan-1(Ry/Rx)

Care must be used in interpreting this last result. A vector with components Rx=-1 and Ry=1 (angle=135o) and the vector with components Rx=1 and Rx=-1 (angle=315o or -45o) have the same ratio Ry/Rx even though they are in different directions.. Your calculator does not which angle you actually want and always give you the smallest angle.
It is up to you to determine which angle to use (that is, either the one given by your calculator or that angle plus 180o).

All angles measured counter-clockwise from EAST.
sin(30o)=.50 cos(30o)=.87
sin(60o)=.87 cos(60o)=.50
sin(45o)=.71 cos(45o)=.71
sin(37o)=.60 cos(37o)=.80
sin(53o)=.80 cos(53o)=.60
sin(0o )=0.0 cos(0o )=1.0
sin(90o)=1.0 cos(90o)=0.0
sin(180o )=0.0 cos(180o )=-1.0
sin(143o)=.60 cos(143o)=-.80
sin(270o)=-1.0 cos(270o)=0.0

Interactive Example

Find R = A + B

Enter the data below, after entering a number click outside of the box each time. If you made a mistake you will be coached. If nothing happens you got it right! Congratulations.

Complete the following table for vector A with magnitude 10 and angle 37 degrees and the vector B with magnitude 20 and angle 143 degrees.
x-componenty-component
A
B
R

Now calculate the magnitude and direction of R and enter your results to the nearest whole number below:
|R|
q

Subtracting Vectors.


To subtract the vector B from A you add the vector -B to A.The vector-B has the same magnitude as B but in the opposite direction as shown below.
The y-component of -B can be calculated in this instance by using either
-|B|*sin135o or |B|*sin315o

Interactive Example

Find R = A - B

Enter the data below, after entering a number click outside of the box each time. If you made a mistake you will be coached. If nothing happens you got it right! Congratulations.

Complete the following table for vector A with magnitude 10 and angle 37 degrees and the vector B with magnitude 20 and angle 143 degrees.
x-componenty-component
A
-B
R

Now calculate the magnitude and direction of R and enter your results to the nearest whole number below:
|R|
q

II. Projectile Motion

Projectile motion is very similar to the free fall problems in Chapter 2. The only difference is that in the case of projectile motion the object is given an additional uniform horizontal motion . This motion is totally independent of the vertical motion. It is this feature that makes these problems easy to solve.

The Connection Between Chapter 2 Problems and Chapter 3 Problems


Recall the following problem from Chapter 2 practice problems:


  

9) How far (to the nearest meter) can a sprinter, running at 9.1 m/s, travel in the time it takes a rock to fall meters?
In the motion picture at the left the green ball (starts out orange) is the falling rock and the red ball (ends up orange) is the sprinter. Ignore the yellow ball for the time being. The button on the left plays the movie while the buttons on the right step backwards and forwards.

Note: if you do not have the quicktime plugin you can view a GIF animation version of the movie here.


  1. You worked this problem by finding the time it took the rock to fall 47 meters, using

    y=yo+vo*t-4.9*t2 ,
    where y=0, vo (the initial vertical speed)=0 and yo=47.

  2. After using this to find the time, t, for the rock to fall, you found the distance the sprinter ran from
    x=xo+vo*t,
    where xo=0, vo=9.1,a=0 (the runner is at a constant speed), and t is the time you found previously.

  3. Suppose the problem is changed slightly to:
    9) How far (to the nearest meter) can a rock, travelling horizontally at 9.1 m/s, travel in the time it takes another rock to fall meters?

  4. I think you can see that nothing essential has changed and that the problem will be worked the same way.

  5. However,in Projectile Motion, the rock moves horizontally as if it were not falling and falls as if it were not moving horizontally (that's what we mean by the motions being independent), there is no need for two rocks in the problem. So we can again reword the problem:
    9) How far (to the nearest meter) can a rock, travelling horizontally at 9.1 m/s, travel in the time it takes this same rock to fall meters?


  6. Again this is the same problem and worked the same way as before except that for the horizontal motion we use vox instead of vo and in the vertical motion we use voy! Now go back to the movie and view it again. This time realize that the yellow ball is the projectile and the green ball is its falling motion and the red ball is its horizontal motion.

  7. What if I ask:
    9) A ball is thrown at an angle of 0 degrees with an initial speed of 9.1m/s from the top of a wall meters high. At some time later it hits the ground. How far does it go?
    (except for the angle of 0 degrees this is one of the Chapter 3 homework problems.)
    This again is the same problem as before and worked the same way!

  8. If you can work the chapter 2 problems, you can work the chapter 3 problems.

General Technique for Solving Chapter 3 Problems


  1. If an initial speed and angle is given find:
    vox=vocosq,
    where vo is the initial speed and q is the angle
    voy=vosinq,
    where vo is the initial speed and q is the angle

    In the above, vox represents the constant horizontal speed and voy the initial vertical speed.
    For an angle of 0o and speed of 9.1m/s this gives us vox=9.1 and voy=0, which makes this last problem statement above the same as the previous one!

  2. Just as in Chapter 2, you need to find out the unknowns:
    xo (the initial position)_
    yo (the initial height) _
    x (the final position)_
    y (the final height)_
    vox (the initial horizontal speed)_
    voy (the initial vertical speed)_
    vx (the final horizontal speed)_
    vy (the final vertical speed)_
    t (the time from beginning to end)_


  3. For the horizontal motion, since there is no horizontal acceleration you have the following two equations (which are the same as the two equations you used for the "sprinter" in problem 9, Chapter 2)
    x=xo+vox*t
    vx=vox

  4. For the vertical motion you can use two of the following equations (which are the same as those you used for the falling rock in problem 9, Chapter 2.)
    vy=voy-g*t
    y=yo+voy*t-.5g*t2
    vy2=voy2-2*g*(y-yo)

  5. Since you have 9 unknowns and at most 4 equations to use, you must find out 5 pieces of information before you can work any problem. Consider the problem which initiated this discussion:


    A ball is thrown at an angle of 0 degrees with an initial speed of 9.1m/s from the top of a wall 46 meters high. At some time later it hits the ground. How far does it go?

    What do we know? We know that vox=9.1 and voy=0 (that is two pieces of information). We also know that (almost always) xo=0 and in this case yo=46 (and that makes four). To work the problem we need one more piece of information. The missing information is simply y=0, the maximum distance travelled by the rock corresponds to the value of x when it hits the ground, that is, when y=0.


  6. additional hints:
    1. many of the hints from Chapter 2 apply here so consult those.
    2. A projectile is at its maximum height when it has stopped moving up. That will tell you that vy=0 at that point.
    3. Just like in the above example, the maximum distance travelled will usually correspond to y=0.


    Interactive Example

    A golf is hit with a speed of 40 m/s at an angle of 37 degrees, what is its maximum height during its flight.

    Enter the data below, after entering a number click outside of the box each time. If you made a mistake you will be coached. If nothing happens you got it right! Congratulations. Use 10 m/s2 for the acceleration of gravity, g.


    xo (the initial position)
    yo (the initial height)
    x (the final position)
    y (the final height)
    vox (the initial horizontal speed)
    voy (the initial vertical speed)
    vx (the final horizontal speed)
    vy (the final vertical speed)
    t (the time from beginning to end)

    Once you know the 5 'givens' (vx is not a given it is calculated from the equation vx=vox) you can use the equations below to solve the problem, but only 2 of the 3 on the right.
    Follow this strategy:

    1. Identify what you are looking for, in this case 'y', and see if there is any equation with it on the left hand side, if all the variables on the right are known, use this equation.

    2. See if any of the other equations containing what you are looking for contain otherwise only known variables. If so use this and algebra to find your answer.

    3. Find any equation containing only one unknown and solve for it. Enter it in the table above and repeat steps 1 2 and 3 as often as needed.

    4. Rarely, this simple approach won't work, and, in that event, you have to solve the equations simultaneously.

    Choose your equation now.
    x=xo+vox*t
    vx=vox
    vy=voy-g*t
    y=yo+voy*t-.5g*t2
    vy2=voy2-2*g*(y-yo)

    Now use the same technique to find the equation for the time, t.

    x=xo+vox*t
    vx=vox
    vy=voy-g*t
    y=yo+voy*t-.5g*t2
    vy2=voy2-2*g*(y-yo)

    Your final task is to find the equation that will find x.

    x=xo+vox*t
    vx=vox
    vy=voy-g*t
    y=yo+voy*t-.5g*t2
    vy2=voy2-2*g*(y-yo)

    © 1999 Carl Adler