Recursos de la 1ª unidad didáctica desarrollada

The study of the free fall motion of bodies near the surface of earth is an important example of the use of the just developed formalism. With“free fall” we mean an idealized situation in which the air resistance can be neglected and the bodies move only under the action of gravity We anticipate that under these conditions the vertical and horizontal motions are independent from one another, as we shall study in Sect.3.7, and that any free body moves with a constant acceleration, g, which is vertically directed downwards and has a magnitude (approximately) g = 9.8 m/s2. We choose a reference frame with the z-axis vertical upward, and the x and y-axes in a horizontal plane, for example the ground. The acceleration of the body, that we shall consider point-like, P, has the components

a¼ 0; 0; gð Þ ¼ gk. ð1:71Þ

The motion of the body depends on the initial conditions. If for example we drop the body from a certain height with null velocity it will move vertically down with uniform acceleration. If we launch it vertically upwards it will gradually slow down, stop and then fall down. If we launch it at an angle with the horizontal it will describe a curved trajectory, etc. Let us study these motions.

Let us start from the simplest case. We drop the body at the height h above ground with null velocity at t = 0. The initial conditions are

x 0ð Þ ¼ 0; y 0ð Þ ¼ 0; z 0ð Þ ¼ h; txð Þ ¼ 0; ty0 ð Þ ¼ 0; tz0 ð Þ ¼ 0:0 The x component of the velocity at the generic time t is

txð Þ ¼ txt ð Þ þ0 Zt

0

axð Þdt ¼ 0 þ 0:t

The x component of the velocity is identically zero (i.e. is zero at every instant of time) because the x components of both acceleration and initial velocity are zero. A similar argument leads immediately to conclude that also x(t) = 0. The same is true for the y components of velocity and position vectors. Notice that the initial conditions x(0) = 0 and y(0) = 0 depend on the reference frame. Its origin has been chosen in such a way to have the point from which we drop the particle on the z-axis. A different choice would have led to the initial conditions, say, x(0) = a, y(0) = b. The two co-ordinates as functions of time would have been x(t) = a, y(t) = b. The motion obviously is the same.

We have found that the motion is along the z-axis. As the acceleration is con- stant, it is uniformly accelerated (acceleration may have both signs, if we want to be

specific we can say accelerated, if the acceleration is positive, delayed if it is negative). Let us nowfind the velocity in the z direction.

tzð Þ ¼ tzt ð Þ þ0 Zt 0 azð Þdt ¼ 0 t Zt 0 gdt¼ gt: ð1:72Þ

Velocity is always negative. Indeed the body moves always in the z direction we have chosen as negative. We now integrate once more to find the position as a function of time z tð Þ ¼ z 0ð Þ þ Zt 0 tzð Þdt ¼ h t Zt 0 gtdt¼ h 1 2gt 2 _ð1:73Þ

which is the law of the motion. Knowing completely the motion, we can look for interesting properties, for example the time taken to reach the ground. This is the instant in which z = 0, hence tf ¼

ffiffiffiffiffiffiffiffiffiffi 2h=g p

and the velocity in that instant tf ¼ t tf ¼pffiffiffiffiffiffiffiffi2gh: ð1:74Þ Consider now the same initial conditions with the difference that the initial velocity has a nonzero vertical valueυ0. With the same arguments as before, we obtain

tzð Þ ¼ t0t  gt; ð1:75Þ

z tð Þ ¼ h þ t0t1 2gt

2_{: ð1:76Þ}

We should now distinguish the two cases of positive (downwards) and negative (upwards) initial velocity.

Ifυ0< 0, the velocity is always negative. Tofind the instant t in which the body is at the height z we solve Eq. (1.76), obtaining

t zð Þ ¼t0 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi t2 0þ 2g h  zð Þ p g :

We have two solutions because Eq. (1.76) is of second degree in t. However, in the case we are considering, one of them, the one with the negative sign, is always negative and consequently does not have physical meaning. We must choose the solution with a positive sign, because the motion starts at t = 0.

The time of arrival at ground, the duration of the fall, is the time at which z = 0, namely tf ¼ t z ¼ 0ð Þ ¼ t0þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffit2 0þ 2gh p g ;

which is shorter than in the case of null initial velocity. Obviously the expressions found in the latter case are particular cases.

Ifυ0> 0, from Eq. (1.75) we see that the velocity is positive, namely upwards, for a while, but it diminishes with increasing time. It is zero in the instant tm=υ0/g, and negative in later times. Indeed, the body reaches the maximum height at tm, namely zm¼ z tmð Þ ¼ h þ t20= 2gð Þ (see Fig. 1.22a). In this case both roots for t(z) have physical meanings provided t≥ 0. Indeed, the body goes twice through the same height, if it is z≥ h, first going up later going down. If z < h one solution is negative and again does not have physical meaning.

Why may it happen that a mathematical solution should be discarded on physical grounds? The reason is that the equations stating the“initial conditions” do not give information on the system before the“initial” instant. In this case the body stood still, say in our hand. But it would have been possible that it was moving upwards in such a way as to reach z = h at t = 0 with velocity equal toυ0. The discarded solution would have made sense.

We now suppose that the initial position is again at the height h above ground, but that the velocity v0is at an angleα with the horizontal. This is what happens when shooting with a cannon from the top of a tower. We choose the z vertical upwards as before and the x horizontal in the plane of z and of the initial velocity. The initial conditions are

x 0ð Þ ¼ y 0ð Þ ¼ 0; z 0ð Þ ¼ h; txð Þ ¼ t00 cosa; tyð Þ ¼ 0; tz0 ð Þ ¼ t00 sina: O x z y v₀ h x_f z_m O z h z_m P _P (a) (b)

Fig. 1.22 Free fall trajectories with initial velocity a vertical upward, b at an angleα with the horizontal

The motion is in the plane xz, as show in Fig. 1.22b. We find, as usual, the velocity using Eq. (1.69) and the initial conditions.

v tð Þ ¼ it0cosa þ k t0ð sina  gtÞ: ð1:77Þ We see that the horizontal, x, component of the velocity is constant and equal to its initial value and that the vertical one, z, decreases linearly in time, exactly as in the case we have considered.

We integrate once more and use the initial conditions to obtain the law of motion,finding r tð Þ ¼ i t0ð cosaÞt þ k t0ð sinaÞt 1 2gt 2_{þ h}   ð1:78Þ or x tð Þ ¼ t0ð cosaÞt; z tð Þ ¼ t0ð sinaÞt 1 2gt 2_{þ h: ð1:79Þ}

We now know completely the motion. If, for example, we want to know the shape of the trajectory we must eliminate t from the equations for the co-ordinates. From the first one we have t ¼ x= t0ð cosaÞ, which, substituted in the second equation, gives

z¼ x tan a  x2 g 2t2

0cos2a

þ h; ð1:80Þ

which is the equation of a parabola. The distance xfat which the body touches the ground, namely the range of the weapon, is the value of x corresponding to z = 0. We then put this value in Eq. (1.80) and solve for x. Wefind

xf ¼ t2 0 g sina cos a 1  ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1þ 2gh t2 0sin 2_a s ! : ð1:81Þ

The negative root solution is for t < 0 and corresponds to the intersection of the parabola on the left of the tower. It is shown dotted in Fig.1.22b and should be discarded. The positive root is the solution for which we searched.

We nowfind the duration of the shot, which is the time tfat which the body touches ground. With x = xfthefirst of the (1.79) solved for t gives

tf ¼ xf t0cosa¼ t0 g sina 1 þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1þ 2gh t2 0sin 2_a s ! : ð1:82Þ

We nowfind the maximum height zmreached by the body. This can be done in different ways. One is noticing that this is the height at which υz = 0. From

Eq. (1.78) we see that a happening at tm¼ t0sina=g, which was substituted in the second Eq. (1.79), gives

zm¼ t2

0sin 2_a 2g þ h:

The same result can be reachedfinding the maximum of the second Eq. (1.79). It is interesting to consider the special caseα = 0. We want the time tftaken by the bullet to reach ground. Equation (1.79) become

x tð Þ ¼ t0t; z tð Þ ¼ 1 2gt

2_{þ h:}

The bullet hits the ground in the instant tf ¼pffiffiffiffiffiffiffiffiffiffi2h=g, which, as we see, is independent ofυ0.This implies that for whatever initial velocity, even if enormous, the time taken to fall from the height h is always the same and is then equal to the free vertical (the special case υ0= 0). In other words, the vertical and horizontal motions are independent.

The law of independence of (the components of) motion was discovered by G. Galilei. In the “Dialogue concerning the two Chief World Systems” he writes (translation by the author):

…suppose having on the top of a tower a horizontally arranged culverin (a relatively light cannon) andfiring point-blank shots, namely parallel to the horizon; then for little or much gunpowder charge given to it, such that the cannonball would fall at a distance of either one thousands arms, or four thousand, or six thousand, or ten thousand, etc., all these shots would take place in times equal to each other, and each equal to the time the ball would take to fall from the cannon’s mouth to earth, when dropped, without any other impulse, for a simple vertical fall. Indeed it looks really wonderful that in the same short time of the vertical fall from a height, for example, of one hundred arms, could the same ball travel either four hundred, or one thousand, or four thousand, or even ten thousand arms, in such a way that in all the point-blank (horizontal) shots it would be in the air for equal times. A little later Galilei specifies that that would be true

…when there were no accidental impediments by the air…