Section 1- 3 - Motion in Two Dimensions

Motion in Two Dimensions
Projectile Motion in Two Dimensions
Shape of a Projectile's Path
The Range Equation
Relative Velocity
Correlation to your Textbook

Motion in Two Dimensions

We looked at two special cases of two dimensional motion, projectile motion and circular motion.

Projectile Motion in Two Dimensions

Projectile motion for us analyzes the motion of a point mass as it travels above the earth’s surface, with only gravity acting on it. We assume that the altitude above the surface is small enough that we can take the acceleration due to gravity to be constant in both magnitude and direction.

We shall make three assumptions here:

Once an object is launched, its acceleration is a constant ag downward.

Air resistance is negligible.

The motions in the horizontal and vertical directions are independent.

This last factsicle we confirmed by observing a demonstration. First, two balls were released from rest at the same time and allowed to fall toward the table; they arrived at the same time. Then, one ball was dropped while the other was launched horizontally from the same height at the same time; once again, they arrived at the same instant. We concluded that the vertical motion was independent of any horizontal motion the objects may have possessed. (In fact, there are some situations where this is not true, for example when the air flow around an object is turbulent, but we'll ignore them.)

So, we may write our kinematic equations separately for the two directions, using components:

The only quantity obviously common to the two systems is the time, t.

EXAMPLE

A ball is rolled horizontally off a table 1.5 m in height at 5 m/s. How far from the base of the table will the ball strike the floor?

Draw a figure to help visualize the situation, including a system of axes with an origin (here, at the edge of the table, with y positive upward and x positive to the right):
Let’s make our inventory:

xi = 0 m	yi = 0 m
xf = ? ←	yf = - 1.5 m
vxi = +5 m/s	vyi = 0
vxf = +5 m/s	vyf = ?
ax = 0	ay = - 10 m/s2
t = ?

Since we are trying to determine something about the x motion, let’s try to find a kinematic equation with only what we know and want to know. Unfortunately, there is no such equation. Here is an ‘eighty percent rule:’ for problems like this one, you’ll probably want to use K Eq 3 for both x and y, and link the motions through the time. If that doesn’t work, then try something else.

Since this is potentially a quadratic equation (we’re looking for the time), let’s put in the numbers now.

Well, turns out it’s not, so let’s solve directly:

Since the ball hits the floor after it left the table at t = 0, we want the positive root.

Now that we have some extra information, we’ll go back to the x-side:

EXAMPLE

Take the same ball as above and launch it at vo = 5 m/s at a 30o angle above the horizontal. How far from the foot of the table will it land? We’ll use the same origin and co-ordinate system as above.

Write down all we know:

xi = 0 m	yi = 0 m
xf = ? ←	yf = - 1.5 m
vxi = vo cosθo = 5 cos(30o) = +4.33 m/s	vyi = vo sinθo = 5 sin(30o) = +2.50 m/s
vxf = +4.33 m/s (no acceleration)	vyf = ?
ax = 0	ay = - 10 m/s2
t = ?

If you compare this with the last example, you see that we have exactly the same bits of information. So, let’s use the same solution method.

Re-arrange to prepare for a quadratic solution,

which is

Once again, we want the solution in the future, +0.85 sec.

Then,

We should probably try a problem when the 80% rule doesn’t work.

EXAMPLE

An object is thrown horizontally from the top of a tall platform of height h. Three seconds later, it hits the ground at a 45o angle. With what speed was the object thrown, and how tall is the platform?

Here, the origin was chosen to be at the foot of the platform and the directions are as shown.

xi = 0 m	yi = h
xf = ?	yf = 0 m
vxi = vxi = vo cosθo	vyi = vxi = vo sinθo
vxf = vxi	vyf = ?
ax = 0	ay = -10 m/s2
t = ?

Believe it or not, there is enough information here. Consider the final velocity and its components:

We see that vo = vxi = vxf = - vyf. So,

As for the height,

Shape of a Projectile's Path

Let's try to determine the type of path (the trajectory) a projectile will take through space near the surface of the earth. That is, find y(x). Start once again with the kinematic equations:

Combine them and eliminate the time. Let’s choose up to be the +y direction, and we’ll start the object’s trip from the origin.

xi = 0	yi = 0
xf = x	yf = y
vxi = vo cosθo	vyi = vxi = vo sinθo
vxf = vxi	vyf = ?
ax = 0	ay = ag
t = ?

Substitute this into the y-equation:

This is a bit of a mess, but what’s important is that, once the object is launched, vo and θo are constant. Their values may change from launch to launch, but once an object is launched, their values are fixed. In that case, we can rewrite the equation above as

with A and B constants. You may recognize this as an example of a parabolic curve. The exact shape will of course depend on vo, θo, and ag, but whenever an object is launched such that a uniform gravitational field is the only agent acting on it, the trajectory will be parabolic.

Later in the semester, we’ll see what happens when the object gets away from the earth’s surface.

The Range Equation

Let's discuss a particularly special case of projectile motion which is of historical interest. In the 17th and 18th century, being a physicist usually meant being an artillery officer. Consider a flat plain (which is also a plane) on which are located a battery and a target. Given an initial projection angle o (the elevation) and speed vo (the muzzle velocity, for guns or cannon), how far will the projectile land from the gun (range, R)?

We'll assume, since the plane is horizontal and flat, that yo = yf = 0.

xi = 0	yi = 0
xf = R	yf = 0
vxi = vo cosθo	vyi = vo sinθo
vxf = vxi	vyf = ?
ax = 0	ay = ag
t = ?

Thjs time, we’ll solve the y-equation for the times that the projectile is on the ground at y = 0 and substitute into the x-equation.

One solution is clearly t = 0. We already knew that, since the object starts on the ground. The other solution is

Rewrite this as a less messy

On the x side,

In the last step, we used a trig identity: 2 sin(α) cos (α) = sin(2α).

EXAMPLE

What angle(s) give(s) the maximum range for a given muzzle velocity? The sine function is a maximum (1) when the angle (2o) is 90o, so the elevation angle should be half that, or 45o, to attain maximum range.

EXAMPLE

Second, for a given range and muzzle velocity, at what angle(s) should a projectile be launched? Let the range be 2000m and the muzzle velocity be 250 m/s. Re-arranging the range equation results in:

It turns out that this is not the only angle that will get the projectile to its target. Remember, we want an angle for which sin(2θo) = R |ag|/vo2. Generally, there are two such angles, as can be seen from the graph. These two angles will be complementary, since the curve shown is symmetric. So, in this example, 90o – 9.3o = 80.7o would also work.

So, what's the difference between these two paths? At a low angle, the projectile is not in the air long, but it has a high x-component of velocity, while at a high angle, the projectile spends a lot of time in the air, but has a correspondingly lower x-component of velocity. These two effects combine to give the same final x displacement as for the low angle case. The first is useful in tank warfare, where it is important to hit the other guy before he gets off a shot at you, while the second is good if there are fortifications around your target. How long are these times? We saw above that the time to reach the target is

and so the time for the other path is

What would be the maximum altitude H reached by the projectile in each case? The time to reach maximum altitude should be half of the travel time to the target, due to the symmetry of the trajectory.

xi = 0	yi = 0
xf = ?	yf = H
vxi = vo cosθo	vyi = vxi = vo sinθo
vxf = vxi	vyf = 0
ax = 0	ay = ag

What can we say about the maximum altitude hmax attained by the projectile?
vy2 = viy2 + 2ay (y - yi)
Let yi = 0, viy = vosino, ay = ag, and let up be positive.
At the highest point in the trajectory, vy = 0 and y = hmax, so,
0 = [vosino]2 + 2ag (hmax)
hmax = [vosino]2/(-2ag) = [vosino]2/2|ag|

What if R |ag|/vo2 >1? Then there is no angle that can get the projectile to the target; it will fall short.

Example: At what angles can a projectile be launched at 300 m/s in order to hit a target on the same level which is 5 km away?

If R |ag|/vo2 = 1, then only 45o will do the job. But if R |ag|/vo2 < 1, then there are two angles which will work:

Relative Velocity

On occasion, it is useful to consider the motion of an object with respect to an origin/co-ordinate system which is itself in motion to some third reference frame. The simple example given in class was that of the moving sidewalk-type 'people mover' seen at most large airports. These conveyances allow the weary to ride from one end of the concourse to the other, while also providing those in a rush a little extra speed as they run down the walkway. For example, consider such a walkway (W) which moves with a speed of +2 m/s with respect to the ground (G):
vW,G = +2 m/s (once again, we'll use the sign of the number to indicate the direction of the vector quantity).
The subscript notation vA,B gives the velocity of A with respect to B.
Now, think of a person (P) walking in the same direction at 1 m/s along the walkway: vP,W = +1 m/s.
It's easy to see that the person's velocity with respect to the ground is +3 m/s, since
vP,G = vP,W + vW,G.
What is the person decided to walk backward on the walkway, such that vP,W = -1 m/s? Then, vP,G would be only +1 m/s; the person would still be going in the same direction as before, although more slowly. If he walked at -2 m/s with respect to the walkway, he'd actually be standing still with respect to the ground, and incidently greatly annoying his fellow travelers.

Question: Suppose that this traveler walks the correct way on the 'people mover' at a speed of 1 m/s, while his twin brother walks at 2 m/s the correct way on the return walkway. What is the relative speed between the brothers?

Now, let's examine the relationship above more carefully:
vP,G = vP,W + vW,G.
One of the hardest aspects of relative velocity is to determine which two quantities get added to obtain the third. Notice, though, that in the example, the lettre 'P' representing the person occupies the first subscript position in both of the terms it appears in, while the 'G' occupies the second position in both terms; only the 'W' changes position. A quick way to remember is to add the two terms from which one subscript changes position.

Since we can add velocity vectors independently in the x, y, and z directions, we can rewrite the relationship above more generally as
vA,C = vA,B + vB,C.

Here's another example:
Consider a boatman who wishes to cross a river (100 metres wide) from one dock north to another exactly opposite. His boat will make 20 m/s in calm water. The velocity of the water is 8 m/s eastward. He aims his boat exactly northward and sets off. How far downstream (x) will he actually land, in what compass direction did he actually travel, and how long will it take him to get there?

We write the relative velocity relationship as
vB,G = vB,W + vW,G, where B = 'boat,' W = 'water,' and 'G = 'ground.'
There are a number of ways to solve this; the shortest may be to realize that, if the velocity components are all constant, a displacement component diagram can be constructed,
rB,G = rB,W +rW,G,
where each term is parallel to the corresponding velocity term. The two triangles so formed are then similar, and so there is a proportionality of the lengths of the sides:
20/8 = 100/x; x = 40m (downstream).
The direction traveled can be found using the tangent of the angle :
tan = 8/20;  = 21.8o.
For the time, we consider that the motion northward (in this case) is independent of the motion eastward; it wouild take 5 seconds to cover 100 metres at 20 m/s.

Now, suppose that, having learned his lesson, he tries again to cross directly to the other side. In what direction should he aim his boat (relative to north) to arrive exactly at the other dock, and how long will it take him??

The common error here is simply to flip the triangle over. But what should be done is to deform the triangle by sliding vW,G over until the sum, vB,G, is pointing due north:
Before, the two short sides of the right triangle were vB,W and vW,G, but now those vectors are the hypotenuse and a short side, respectively. So,
' = arcsin(8/20) and |vB,G | = 20cos' Finish the calculations yourself.

Mastery Question

Suppose that a man tosses a ball vertically while riding a train moving at constant velocity vT. What shape trajectory will the man see the ball take? What shape trajectory will someone standing by the side of the track see the ball take?
Now, suppose that the train accelerates forward (aT) and the ball is launched as before when the train has velocity vT. What path will the observer in the train see the ball take? What path will the stationary observer see?

Click here for solution.

Try this Applet: http://users.erols.com/renau/velocity_composition.html

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What if the angle had been other than 45o? Then, vxf = vyf tan θf and proceed as above.