CHAPTER 2

MOTION IN TWO DIMENSIONS

 

A. Projectile Motion
B. Range of a Projectile
Review Problems for Chapter 2
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In Chapter 1 we examined objects which moved only in the horizontal or vertical direction.   In particular, when we considered the motion of a freely falling body, our discussion was limited to examples where the body only moved up and down.  That is cases where we remove this restriction and investigate a more general type of motion called projectile motion.

Before beginning this chapter it is advisable that you review the appendix dealing with trigonometry (Appendix E – Trigonometry, page A-21).

 

A.       Projectile Motion

 

A simple example of projectile motion is a ball rolling off a table.  Let us define the horizontal direction as the x-axis and the vertical direction as the y-axis.  Consider a ball initially rolling on off a flat table with an initial velocity of 5 m/sec.

                               

While the ball is on the table we observe that:

                The initial x-component of velocity = vIx = 5 m/sec = Constant

                The initial y-component of velocity = vIy = 0 = Constant

                The x-component of acceleration = ax = 0 = Constant

                The y-component of acceleration = ay = 0 = Constant

When we refer to the components of acceleration and velocity we mean that part of the velocity or acceleration that points in the x or y direction.

Now consider what happens the instant the ball leaves the table.  The initial velocity in the y-direction is still zero and the initial velocity in the x-direction remains 5 m/sec.  Now, however, the ball is no longer in contact with the table and it falls freely.   The gravitational acceleration of the ball is down.   Examine the motions in the horizontal and vertical directions independently.

Horizontal motion :

Since there is no acceleration in the horizontal direction the x-component of velocity is constant.   So we have,

                          ax = 0

                          Vx = Constant

                          x = vIx t                                                                                               (2.1)

Vertical Motion :

In the vertical direction there is an acceleration equal to the acceleration of gravity.   Thus the velocity in the vertical direction changes and we have, 

                                   ay = g

                                                                                           (2.2a,b,c,d)

Note :       The only variable common to both horizontal and vertical motion is time.

Examples:

1.             An outfielder throws a ball from right field with a velocity of 132 ft/sec (90 mph).   If the ball is released from a height of 6 ft with only horizontal velocity, how far does it go before hitting the ground?

Let us begin by drawing a picture of this problem.

Now list the information we are told and can infer:

                                                Horizontal                                               Vertical

                                                                          

Remember from Chapter 1 that we can only solve a kinematics problem of we know at least three variables.   With this in mind we can see hoe we must proceed to solve this problem.  We can use this information provided in the vertical direction to find hoe long the ball is in the air.  This gives us three variables in the horizontal direction (remember time is common to both) and we can determine x.  So first use Eq. 2.2c to find time.

                                                                            

                               

Now that we know that the ball is in the sir for 0.61 sec we can determine the horizontal distance it has traveled using Eq. 2.1

                               

Before moving on to the next example consider the implications of this problem.  Even though the ball was thrown with a very high velocity of 90 mph it only traveled 80.5 ft.   Compare this with the 90 ft distance between the bases in baseball.   It is clear that we must find a way to throw the ball in the air a distance further than 80.5 ft.  Since it is difficult to throw the ball faster than 90 mph, a longer throw must be accomplished by keeping the ball in the air for a longer time.   This is achieved by throwing the ball in an arc.   When we do this, the ball will have an initial vertical velocity.

2.             As a second example consider a more general problem where the projectile has an

initial velocity in both the horizontal and vertical directions.

A ball is thrown with a velocity of 30 m/sec (66 mph) at an angle of 30° above the horizontal.

a.         How long does it take to reach its highest height?

b.        How high is this?

c.         What is the total time the ball is in the air?

d.        How far in the horizontal direction has it gone?

First draw the picture.

Now determine the components of initial velocity.

List the information we are told and can infer:

                Horizontal                                              Vertical

                                                                             

a.  Use Eq. 2.2a to find the time to reach this highest point.

                                               

b.        Find the maximum height by using Eq 2.2b

                                                Therefore, the highest height h = 11.25 m                       

c.         The time for the ball to reach its maximum height is equal to the time for the ball to fall from its maximum height.  Therefore the ball is in the air for a time

d.        Use Eq 2.1 to find how far in the horizontal direction the ball has gone.

Compare this result, d = 78 m = 256 ft, with the distance we found in Example 1. 

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B.      Range of a Projectile

 

                The total horizontal distance a ball travels while it is in the air is called to range if the projectile.

If T = time of flight if the projectile, and vIx = v I cos θ is the horizontal component of velocity, then the range of the projectile is given by Eq 2.1 as

                               

From this equation it is apparent that if you wish to change the range of the projectile, you can do so by altering the initial velocity, the angle of release, or the time of flight.  Since the time of flight is determined only by the vertical motion let us perform some simple algebra to eliminate T from the range equation.   Do this by determining the time it takes for a ball to reach its maximum height in the problem below.

 

                Horizontal                                              Vertical

                                                  

Eq 2.2a can now be used to find the time to reach the maximum height.

                               

Therefore,        

The time to reach the maximum height is half the total time of flight.   So we have

                               

Now substitute this into the range equation

                               

                                                                                                     (2.3)

We now have an expression for the range of a projectile in terms of the initial velocity, the initial angle of projection, and the acceleration of gravity.  This equation can be further simplified by making use of a simple trigonometric identity.

                               

Now Eq 2.3 may be written as

                                                                                                                     (2.4)

We can see from this reaction that the range of a projectile can be increased by increasing the initial velocity or by decreasing the acceleration of gravity.  This equation also tells use what initial angle of projection will give the maximum range.   The maximum value of the sin function is one.   This occurs when the argument of the sin function is 90°.   Thus the maximum value of sin2θ is when θ = 45°.

Example:                 Estimate the maximum distance a long jumper can jump.   Determine some reasonable values to use in this problem.

First assume vI = 30 ft/sec (a world class sprinter).   Let θ = 45° to maximize the range.   Finally use g = 32 ft/sec2.   The range equation gives

                               

                                               

 

                                                Recall that the sin90 ° = 1.  So we have

 

                                               

Question:  This result is very close to the world record jump of 29 feet 2 ½ inches by Bob Beamon in the Mexico Olympics in 1968.   Can you think of any refinements to our initial assumption that will alter our result?

Summary

Projectile Motion                                                                 Vertical Motion

Components of Velocity                                                     Time of Flight

Horizontal Motion                                                               Range

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Review Problems for Chapter 2

 

  1. A marble rolls off the edge of a horizontal table 1.25 meters high.   If the ball is rolling on the table with a speed of 4 m/sec, how far from the end of the table does it land?

  2. A garden hose shoots water horizontally from the top of a tall building toward the wall of a second building 20 meters away.  If the speed with which the water leaves the hose is 5 m/sec, how long does it take the water to reach the second building, and what distance does the water fall in this time?

  3. A bullet is fired horizontally from a gun with an initial velocity if 2000 ft/sec. If the bullet travels a horizontal distance of 1000 ft what vertical distance does it drop?

  4. In the first chapter you were asked to find how long it takes a 90 mph fastball to travel 60 feet from the pitcher’s hand to home plate.   The answer to this question is approximately 0.45 sec. Assume a pitcher releases the ball with horizontal velocity only.   How far does the ball drop in:
    1. the first 0.225 seconds?
    2. the total time of 0.45 seconds?
    3. the final 0.225 seconds?
  5. For the three cases shown below determine the initial horizontal and vertical velocities.

  6. A high jumper takes off with a velocity of 7 meter/sec at an angle of 70 to the horizontal.
    1. What is the jumper’s initial horizontal velocity?
    2. What is the jumper’s initial vertical velocity?
    3. What height does the jumper achieve?  Calculate this distance in meters and feet.

  7. A ball is thrown with a speed of 50 m/sec (110 mph) at an angle of 37 degrees above the horizontal. 
    1. What are the initial components of velocity?
    2. What is the maximum height of the ball?
    3. How long does it take the ball to reach its maximum height?
    4. What is the velocity if the ball at its peak height?
    5. What is the acceleration of the ball at its peak height?
    6. How long does it take the ball to go the entire range?
    7. What is the range of the ball?
    8. Sketch the trajectory of the ball including the components of velocity at various times of flight.

  8. A very good golfer can drive a golf ball with an initial velocity of 150 mph (220 ft/sec).   If the face of a driver is inclined at an angle of 10 degrees, how far would the golf ball fly?  Express your answer in feet and yards?

  9. A typical velocity for a golf ball hit with an iron is 150 ft/sec (102 mph).   Calculate the maximum height of a golf ball with an initial velocity if 150 ft/sec for the following cases:
    1. Golf ball hit with a 1 iron.  The face of a 1 iron is inclined at an angle of 20 degrees.
    2. Golfball hit with a 9 iron.  The face if a 9 iron is inclined at an angle of 45 degrees.

  10. Suppose a baseball player hits a ball so that its take-off angle is 25 degrees.   If the ball travels 350 ft, what was the take-off velocity?

  11. The typical take-off angle for long jumpers is 20°.   Assuming Bob Beamon set his world record of 29 feet, 2 ½ in. with a take-off angle of 20°, what was his take-off velocity?

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