Kinematics in Two Dimensions; Vectors

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I. Vectors

Adding Vectors

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. Note: All angles are measured counter-clockwise from EAST.

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-component	y-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.

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.

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-component	y-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 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 moves horizontally as if it were not falling and it falls as if it were not moving horizontally. It is this feature that makes these problems solvable.

The Connection Between Chapter 2 Problems and Chapter 3 Problems

9) How far (to the nearest meter) can a sprinter, running at m/s, travel in the time it takes a rock to fall meters?

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 46 meters, using
   
   y = y_o + v_ot - 4.9t² ,
   where y(the final height) = 0, v_o (the initial vertical speed) = 0 and y_o(the initial height) = 46 m.

2. After using the above equation to find the time, t, for the rock to fall, you found the distance the sprinter ran from
   x = x_o + v_ot,
   where x_o(the initial horizontal position) = 0, v_o(the initial speed) = 9.1 m/s, 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 m/s, travel in the time it takes another rock to fall meters?

4. You can see that nothing essential has changed and that the problem is worked the same way.

5. In Projectile Motion, an object moves horizontally as if it were not falling and falls as if it were not moving horizontally, thus, there is no need for two rocks in the previous problem statement. One rock projected horizontally from an initial vertical height of 46 m will do the same . The problem statement can be reworded:

   9) How far (to the nearest meter) can a rock, travelling horizontally at 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 the ones before. Now go back to the movie and view it again. This time realize that the yellow ball is the projectile, the green ball is its falling motion and the red ball is its horizontal motion.

7. A projectile launched with a speed v_o at an angle of q degrees has an initial horizontal velocity (called v_ox) of v_ocos(q) and a vertical velocity (called v_oy) of v_osin(q). The problem statement above can now be reworded again:
   

   9) A rock is thrown at an angle of 0 degrees with an initial speed of m/s from the top of a wall meters high. At some time later it hits the ground. How far does it travel horizontally?

   This is one of the Chapter 3 practice problems. As was demonstrated above, this problem is actually a reworded problem from Chapter 2. All the problems in Chapter 3 are similar extensions of Chapter 2 problems.

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,

   voy = vosinq, where vo is the initial speed and q is the angle

   
   In the above, v_ox represents the constant horizontal speed and v_oy the initial vertical speed.
   For an angle of 0^o and speed of 9.1 m/s this gives us v_ox = 9.1 m/s and v_oy = 0.

2. Just as in Chapter 2, you need to find out the unknown variable values:

   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 used for the "sprinter" in problem 9, Chapter 2):

   x = xo + voxt

   vx = vox

   
   
4. For the vertical motion you can use any two of the following equations (which are the same as those used for the falling rock in problem 9, Chapter 2.)

   vy = voy - gt

   y = yo + voyt - 0.5gt2

   vy2 = voy2 - 2g(y-yo)

5. Since you have 9 unknowns and at most 4 equations to use, you must find 5 pieces of information (variable values) before you can work any problem. Consider the problem discussed above:
   
   
   

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

   
   
   What do you know about this problem? For starters you know that v_ox = 9.1 m/s and v_oy = 0 (that is two pieces of information). You also know that (almost always) the initial horizontal position, x_o = 0 and the initial height, y_o = 46 m (and that makes four variable values). To work the problem you need one more variable value. The missing information is y = 0, the maximum horizontal distance (x) travelled by the rock corresponds to the value of x when the rock hits the ground, that is, when y=0. See the illustration at the right.
   
   
   
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, this corresponds to v_y=0 .
   3. Just like in the above example, the maximum distance travelled by a projectile will usually correspond to y=0.
   
   
   

   Interactive Example

   A golf ball is hit with a speed of 40 m/s at an angle of 37 degrees. a) What is its maximum height during its flight? b) How long does it take to reach this height? c)How far does it travel horizontally during this time?

   As a first step find the values for the five known variable values and enter them below . After entering each value click outside of the box . If you made a mistake you will be coached.

   
   

   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 five 'givens', that is, the five known variable values, (v_x is not a given it is calculated from the equation v_x=v_ox) you can use the equations below to solve the problem, remember that only any 2 of the 3 equations on the right can be used.
   Follow this strategy:

   1. Identify what you are looking for 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. If not...
      
      
   2. See if any of the other equations containing what you are looking for contain otherwise only known variables. If so use this equation and algebra to find your answer. If not...
      
      
   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. In rare cases the straightforward approach outlined above won't work, and, in that event, you have to solve two of the equations simultaneously to obtain the desired results.

   Choose your equation for finding y now:

   x=xo+vox*t vx=vox

   vy=voy-g*t y=yo+voy*t-0.5g*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 select 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)

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   © 1997-2000 Carl Adler