Motion Under Constant Acceleration 2D Motion

We will be studying projectile motion both in the atrium and in Applet land. For now, the following two graphs summarize the kinds of observations you can make about projectile motion:

Relevant parameters:

• Total horizontal range
• Projectile angle
• Time of flight
• Maximum height of projectile
• Time it takes to get to maximum height
• Initial velocity

Out task now is to see how these parameters can be predicted by knowing the initial velocity and angle of a projectile.

Let's review the data that we acquired in the atrium:

Some questions:

1. What are the X and Y components of the velocity?


Determining the X and Y components is fairly simple as shown in the above diagram. Any velocity vector can be deconvolved into its X and Y components by using sine and cosine functions on the angle, A.

For instance, the the angle A was ninety degrees then there would be no X-component to the velocity. Since the cosine of 90 degrees is 0, then V_x cosine 90 = 0 and there is no X component to the velocity.

Similarly, if the angle is 0, then V_y cosine 0 = 0 and there would be no Y component to the velocity. Any other angle would have both X and Y components as defined in the figure above and again below:

• V_x = V cosine A
• V_y = V sine A

2. Why does the maximum distance occur at 45 degrees?

One of the students in class supplied the correct answer by staying that this is the angle in which the product V_x * V_y is a maximum.

This conclusion was reached by understanding, from the atrium experiment, that both the X and Y components of the velocity are important.

How did we know this? We knew this from the 30 vs 60 degree experiment. We established that V_x in the 30 degree case was larger than V_x in the 60 degree case.

Because the 60 degree case achieved a higher maximum height, we also established that V_y was higher in the 60 degree case than the 30 degree case.

But, both cases produced the same horizontal range which indicates that both V_x and V_y are important in determining the total range.

The reason that 45 degrees gives the largest horizontal range is because the product:

cos A * sin A has a maximum value at A = 45 degrees.

3. Why does 30 and 60 degrees give the same horizontal range?

Another student correctly pointed out that this is merely a consequence that cosine 30 = sine 60 and sine 30 = cosine 60 and so the two cases are symmetric.

4. How can we determine the total time of flight?

If you shoot the cannon below you will see a parabolic trajectory. Defining the motion in the two coordinates allows us to say the following:

In the X-coordinate the shot starts from the cannon, call this x_o and then hits the ground at some other distance, call this x_f .

The total time it takes for the projectile to cover this distance in X only depends on the Velocity in the X-direction. Hence the time of flight can be expressed as:

t = (x_f - x_o)/V_x

Now let's consider just the Y-motion. In the Y-direction, the projectile reaches some maximum height, h, and then returns to the ground. So the time it takes to undergo this motion is

t = 4h/V_y

Now note the following: V_y, unlike V_x, is not a constant, but is continually changing. When the object is at maximum height, its velocity is, of course V_y = 0. So there is some average velocity:

1/2(V_y) that holds for the motion of the projective from height = 0 to height = h. Since the object has to go up and down, its total displacement is 2h but its average velocity during this total displacement is 1/2(V_y).

So the travel time t is

t = (2h)/(1/2)(V_y) = 4h/V_y

5. What determines the maximum height of the projectile.?

Well, the students correctly inferred that the maximum height depends on the initial velocity in the y-direction as well as inversely on the gravitational acceleration, g

So that leads to something like:

h = V_y/g

But if we look at the units of that we see that its:

meters = (meters/sec) / (meters/sec²)

but the right had side has units of seconds. Hence, dimensionally we need to multiply the right hand side by meters/second to get meters. This means that we must have something like:

h = V_y/g * V

which would now be dimensionally correct. But what's the physics for this?

Next we did an in class demo about free fall and the measurement of velocities. That lead us to the general result that free fall represents a case of gravitational potential energy converted into kinetic energy. This is also the point of the second lab activity.

Since energy is mostly conserved in free fall (little energy is lost due to air friction). We can write down the conservation of energy between potential energy and kinetic energy or:

1/2 mv² = mgh

or

v² = 2gh

or

h = v²/2g

We see that the mass drops out entirely which, of course, is why two objects of different mass fall at the same rate. If heavier masses hit the ground prior to light masses when dropped at the same time, there would be a violation of energy conservation.

So what determines the maximum height is the initial kinetic energy of the projectile and the angle of launch.

6. What determines the total range?

Now we are in an position to understand what the total range of the projectile is:

Return to the earlier time of flight analysis where we established that:

t = (x_f - x_o)/V_x = 4h/V_y

But we just derived the expression h = V_y²/2g

We can then substitute this expression into the time of flight equality to get:

t = (x_f - x_o)/V_x = (4/V_y)* V_y²/2g = 2V_y/g

Letting R = range = x_f - x_o then leads us to the final expression:

R = (2V_xV_y)/g

or

R = (2V²cosine A sine A)/g

Using the cannon below you can verify this expression directly by measuring the distance using the cursor readout feature now enabled.

To summarize, we were able to derive the correct expression for range of the projectile by using a time of flight analysis in combination with convervation of energy.

7. How does energy conservation help to determine total range and maximum height?

Consider this as a question for the final exam.

The cannon will help: