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Horizontal and Vertical Motion

This section has no new equations but it does present one new idea, so prepare your imagination. A pencil and paper may prove handy.

Imagine you are sitting at the shore of a bay, and you see a boat traveling along at constant speed in a straight line. A sailor at the top of a vertical mast drops a grapefruit. (See Figure 4-6.)
We will pretend air resistance plays no role (mostly true). Where will the grapefruit land?
  1. In front of the mast.
  2. At the foot of the mast.
  3. Behind the mast.
Choose your answer before you read any further.

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Figure 4-6


Few people choose A. Not many people choose B. If you are like most people, you chose C, thinking that somehow the boat moves out from under the grapefruit. If that was your answer, then you need to do some rethinking. Here is what really happens. Figure 4-7 shows two ships at four successive times, one ship at rest and the other in uniform motion. Sailors at the tops of the masts drop grapefruits at the same time. Both grapefruits drop to the foot of the corresponding mast.

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Figure 4-7

What is going on in the previous example? Just after the grapefruit is released from the hand on the second ship, it still has its horizontal motion. If air resistance does not affect it, then it maintains its same horizontal motion from start to finish. The vertical motion, on the other hand, proceeds on schedule regardless of the horizontal motion. At t = 0.2 s, both grapefruits have moved vertically 0.05 m (the second has moved horizontally as well). As time goes on, the second grapefruit keeps up with the ship, and both grapefruits hit the deck at the same time.

If you do not believe the figure, try the experiment yourself. While walking at a constant speed, release an apple above your head (and a little forward). It will fall in front of your face and land at your feet. (See Figure 4-8.)

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Figure 4-8

The picture of a boat in the bay and the resulting principle are due to Galileo:
Horizontal and vertical motion are independent. That is, motion in the x- and y-direction can be considered independently.

Now let's leave the bay and travel to the edge of a cliff with a large plain at the bottom. We have two cannonballs, and at the same time, we shoot one horizontally off the cliff and drop the other. Which will hit the plain first?

Again, most people will say the dropped ball hits first. As long as air resistance plays at most a small role, they will hit at the same time. (See Figure 4-9.) For the second ball, the vertical motion of falling is not affected at all by its horizontal motion. It may help your intuition to realize that the shot cannonball does have a larger total velocity all the way down. Notice that Figure 4-9 looks just like Figure 4-7 with the boats removed. It is the same physical principle.

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Figure 4-9

Up until now we have been using v1, v2, and a to denote velocities and acceleration in one dimension. For these problems we will need to keep track of the vertical and horizontal pieces separately, so we need the symbols ϖ1x, v2x, ax, v1y, v2y, and ay.

The following box shows how this principle gets translated into equations:
If the net vertical force is (Fnet)y. then we can determine the vertical motion using


Similarly, we can determine the horizontal motion using similar equations with y replaced by .v.


Objects in free fall experience only the force of gravity, so we can say more:

For an object in free fall at the surface of the Earth, we have




where "up” is positive and we use the estimate 10.


A cliff stands 80 m above a flat plane. One cannonball is dropped, and another is fired horizontally at 120 m/s at the same time. How far from the first ball will the second ball land?



The first ball falls straight down, of course. Let's DRAW A DIAGRAM for the second ball while it is in flight (Figure 4-10). Note that the cannon exerts a force on the cannonball while it is in the cannon, but after the ball leaves the cannon, the only force is gravity.


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Figure 4-10


We record the information we have. We do not know the mass of the ball, but from Section B we know we do not need it. The acceleration vector is 10 m/s2, down.


Vertical horizontal
v1y = 0 v1x = 120 m/s
Δy = –80 m Δx = ?
ay = –10 m/s2 ax = 0 m/s2
Δy = v1yΔt + 1/2 ayΔt2  
–80 m = 1/2 (–10 m/s2)Δt2  
Δt = 4 s Δt = 4 s
  Δx = v1xΔt + 1/2 axΔt2
  Δx = (120 m/s)(4 s)
  Δx = 480 m

We solved the vertical problem first because we had more vertical information than horizontal information. The time Δt = 4 s was the connection between the horizontal and vertical parts. You should work through this example yourself without looking at the book.
The following example involves a projectile.

A cannon is fired on level ground, so that the ball's initial velocity is 300 m/s and directed 30˚ up from the horizontal. How far from the cannon will the ball fall? (See figure 4-11.)
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Figure 4-11
Interruption: We need to know the horizontal and vertical components of the initial velocity v1. We need to find a horizontal vector v1x and a vertical vector v1y so that their sum is the original vector v1 (see Figure 4-12). We can find the magnitudes of v1x and v1y using simple trigonometry. (You may need to review trigonometry at this point.)

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Figure 4-12

The force diagram is the same as in Figure 4-10. The cannonball rises and then falls to the same height from which it started, so we have Δy = 0 m.
vertical horizontal
Δy = 0 m Δx = ?
v1y = 150 m/s
ay = –10 m/s2 ax = 0 m/s2
Δy = v1yΔt + 1/2 ayΔt2  
0 = v1yΔt + 1/2 ayΔt2  
0 = v1y + 1/2 ayΔt


In the last line, we have divided by a factor of Δt.
0 = (150 m/s) + 1/2 (–10 m/s2) Δt
Δt = 30 s Δt = 30 s
  Δx = v1xΔt + 1/2 axΔt2
  Δx = 7800 m


​We have been talking about grapefruits and cannonballs so far. Objects with a more complicated shape obey the same rules, as long as we use the center of mass to talk about the position of the object. Figure 4-13 shows a baseball bat fired from a cannon. The center of mass moves in a parabola, just like the cannonball in the previous example, even though the bat is rotating. In fact this is a definition of the center of mass. If an object is set to freely rotating, the center of mass is the point which refuses to rotate. The gravitational force acts as if it were exerted only at the center of mass.

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Figure 4-13

In this chapter we looked at the law of gravitation, whose grand form is
Fgrav = Gm1m2/d 2
For most problems near the surface of the Earth we can use simply
Fgrav = mg
where g = GMEarth/REarth 2 = 10m/s2
When any object near the Earth's surface has only gravity acting on it (freefall), it has a downward acceleration vector of magnitude 10 m/s2. This curious result comes from the fact that the force of gravity is proportional to the mass, while the second law of motion states that the acceleration in inversely proportional to the mass.

We also explored the principle that horizontal and vertical motion are independent. This allows us to solve problems involving projectiles, that is, objects with only gravity acting on them. We will see more of this principle in the following chapter.

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