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Free Fall

An object is said to be in free fall when nothing is touching it, so that the only force on it is gravity. Such an object is called a projectile. The simplest problem in free fall involves dropping objects near the surface of the Earth. We want to know which falls faster, a heavy object or a light one? Things become complicated if the object is too light, like a leaf fluttering to the ground, so at first we will consider two objects for which air resistance is only a small consideration. (We discuss air resistance in Chapter 6.) Let us start by doing a pair of examples.



How long does it take a small rock (0.02 kg) to fall from rest 2 meters to the ground?



First, we DRAW A DIAGRAM (Figure 4-3). There is only one force since nothing touches the rock and we are neglecting air resistance. Second, we find the net force


Fnet = Fgrav = –mg


= –(0.02 kg)(10 m/s2)


= –0.2 N


where the negative sign reminds us that gravity points down.


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Figure 4-3


Third, we obtain acceleration by writing

Fnet = ma
a = Fnet/m = –0.2 N/0.02 kg = –10 m/s2


(This result should look familiar.)
We also have


v1 = 0


Δy = –2 m


so that


Δy = v1Δt + 1/2 ayΔt2
–2 m =1/2 (–10m/s2t2
Δt = 0.63 s


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Figure 4-4




How long does it take a medium-sized rock (0.2 kg) to fall 2 meters from rest?



The rock is larger, and so is the force of gravity (and the force arrow, see Figure 4-4). We write


Fnet = Fgrav = –Mg = –(0.2 kg)(10 m/s2) = –2 N

The acceleration is


a = Fnet/M = –2 N/0.2 kg = –10 m/s2

The rest of the problem is the same, so Δt = 0.63 s.

WHAT HAPPENED? The force of gravity was larger in Example 2, BUT the acceleration is inversely proportional to mass in the second law of motion. The net effect is that the acceleration is the same 9.8 m/s2 for both rocks. Try this at home with a pen and a stapler or some such thing. It is difficult to gain an intuitive grasp of this situation, but think about it until you also understand why the two rocks have the same acceleration.

Let us revisit Example 2 in Section 2.H. There we saw an apple tossed straight up. It rose, came to a stop, and fell. We claimed that the acceleration was always negative (that is, down), even at the top point. We are now at a better position to understand why. Once the apple leaves the hand, there is only one force on the apple, the downward force of gravity. (See Figure 4-5.) The acceleration must be down as well. In fact, we now know the acceleration is a constant 9.8 m/s2, down.

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Figure 4-5

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