Firstly, it will be shown that two opposite movements, one uniform and one accelerated, can be present and be exerted at the same time in the same body.

This is obvious. If a seaman moves from the bows to the stern of a boat in an uniformly accelerated movement and meanwhile the boat is carried by the stream of the river in an opposite uniform movement, clearly the seaman is impelled by the uniform movement together with the boat in which he is present and simultaneously he moves by himself in an opposite accelerated movement. Therefore, not only opposite motive forces active at the same time in the same body of the seaman are present, but, moreover, their actions, i.e. opposite movements without percussion are unquestionably exerted in the seaman. However, it is true that this combination of opposite movements results in a displacement in the space of the world either slowed down or nullified. But this does not hinder the existence of opposite movements. Let us thus imagine that the boat is projected upwards in an uniform movement while a stone falls from the top of a mast to the bottom of the boat in an accelerated movement. Is it not obvious that the stone is agitated and propelled upwards together with the boat in a movement of projection and simultaneously falls by its normal gravity? And what is the difference whether the stone is thrown upwards with the boat or without the boat? In either instance it is of course provided with the two opposite motive forces by which it is agitated or impelled without percussion. The truth of this proposition was confirmed by the experiment recorded in chapter 23. The three balls R, R, R (Table 23.​1, Fig. 71) projected from A, D and B with the same degree of impetus with which the whole wheel ABCD rotates must travel equal distances in the same time (the obstacle of air being discarded). The lengths of their falls in the same time are also equal. In the downwards movement BK two impulses, that of projection and that of gravity, striving in the same direction, result in a displacement BK equal to the sum of the horizontal displacement AI and the downwards movement IH, or FH starting from immobility. Consequently, since the upwards displacement DG is equal to the difference of the movements AI and FM, it appears obvious that both movements are exerted at D, i.e. the ascent due to the projection DO equal to AI or BL, and the fall OG equal to FM. Q.E.D.

A heavy body projected upwards by a percussive force will necessarily be raised upwards a certain distance.

A heavy body A is projected upwards perpendicularly to the horizon BA by a percussive force which, however small it is imagined, always impresses in the body A a certain impetus which is not indivisible (Table 23.​1, Fig. 86). This impetus AB in the course of the time AC creates a rectangular surface of the impetus BC which corresponds to the travelled distance. From the instant of the percussion, the natural downwards force of gravity, from an indivisible point, increases the degree of the impetus by equal increments in all the successive instants of the time AC. Therefore, if in the time AC the impetus resulting from the natural gravity by flowing and growing creates the triangular surface of the impetus ADC with straight sides, this will correspond to the distance of its fall while the heavy body is moved upwards by the impetus BA as if it were by a boat. The initial degrees of impetus generated by the force of gravity are always smaller than the impetus of projection BA until, at the end of time AC, the body acquires the maximum degree of velocity DC equal to the velocity AB. Consequently, during all the time AC, a greater impetus BA impels the body A upwards more violently and travels more distance upwards than it is repelled downwards by the impetus of gravity. Therefore, the impetus of projection BA overcomes the resistance of the impetus of gravity by the excess of the velocity of projection BA which always remains of the same degree over the growing impulses of gravity. The body A thus rises until the impetus of gravity by increasing up to DC matches the impetus BA, i.e. until it is pushed upwards by as much as it recedes downwards. Q.E.D.

If a heavy body is impelled vertically upwards in an uniform movement so that in a given time it can travel a well-determined distance, at midpoint of its ascent, the raising of the projected body stops.

A heavy body A is impelled vertically upwards in an uniform movement with an impetus D so that it can travel the distance GF in a time T (Table 23.​1, Fig. 87). I claim that its ascent ends at the midpoint C. Opposite motive forces do not annihilate each other but they remain effective and continue producing their particular effects. An uniform motive force of gravity impels the body A with the impetus D as if there was no opposite force of gravity since it is exerted without percussion. The accelerated motive force of gravity also impels the body as if there was no obstacle. Consequently, the impetus of projection as well as the impetus of gravity, carries out its own forwards movement. It is not different when a seaman runs in an accelerated movement from the bows to the stern of a boat carried by the opposite stream of a river. Actually, in relation to the space of the world, they do not attain their goal. The body A indeed moves upwards in a slowed down movement as a result of the excesses of the stronger impetus D over the very small initial impulses of gravity. Finally, it is forced to end its ascent when the impetus of gravity E by successively increasing matches the impetus D. This will be shown to occur at the midpoint C. FG is prolonged below G. GH is taken equal to FC. The heavy body A falling from its position of immobility G in an accelerated movement travels in the time T the distance GH at the end of which it acquires a degree of velocity E. The same body A falls from point G provided with a double impetus, an uniform impetus with which it can rise up to F in the time T, and simultaneously an opposite and accelerated impetus with which it can travel the distance GH. In such instance, the point from which the fall starts, G, is not stable and does not remain in the same position but is displaced in an uniform movement to F. Meanwhile, the heavy body A by falling arrives at C. Therefore, the impetus E acquired when rising in an accelerated movement or when falling during the same time is exactly the same. If the body proceeded with its displacement in an uniform and even movement, it could travel a distance twice the distance CF thus travelled in an uniformly accelerated movement. GF is twice CF. Therefore, the impetus E is equal to the impetus D since both can travel the same distance GF in an uniform movement in the same time. Actually, opposite impulses when brought to equality produce apparent immobility and the end of the movement in the space of the world. Consequently, rising stops at the midpoint C and does not proceed further. Q.E.D.

In these two propositions, as was done in the others, the obstacle of air which amazingly disturbs the progress of the ball in its ascent and in its descent, and weakens its impetus, must be discarded. This is obviously confirmed by the following experiment. An arrow or a wooden ball thrown vertically by a ballista or a catapult up to a considerable height travels in equal times a longer distance, almost twice as long, when rising than it does when falling. Moreover, its impetus at the beginning of the ascent is greater than the impetus acquired at the end of its fall. This of course could not occur if the impetus of projection did not weaken continuously in the ascent and if, in the descent, the impetus of gravity did not grow by diminishing increments and was not finally reduced to being uniform.

A ball A is thrown vertically upwards along AC with an impetus BA (Table 23.​1, Fig. 86). Unquestionably, in the ascent, this impetus of projection is continuously diminished by the resistance of air so that, at the end of the time AE, EF is equal to the maximum degree of the impetus of gravity progressively growing. Obviously, the trapezium surface of the impetus ABFE is generated by the decreasing impetus of projection. Since the triangular surface of the impetus AEF corresponds to the descent or the resistance of the heavy body A while it is propelled upwards, the triangular surface AFB will correspond to the length of the ascent travelled in the time AE, when the projectile is brought to the upper point of its ascent. In the following time EC equal to EA, the ball falls downwards. Two impulses are then also exerted. One is an impetus of projection upwards in an uniform movement due to the impetus FE, which will generate the rectangular surface of the impetus GE, because, while the ball falls, the consistency of air does neither hinder nor slow down the impetus impelling upwards. But, in the meantime, the natural impetus of gravity creates the trapezium surface of the impetus EFDC in the time EC. All this surface does certainly not correspond to the length of the fall since the opposite impetus of projection subtracts from it the rectangular surface of the impetus GE. Therefore, the surface of the impetus which is exerted in the falling ball will be the triangle FDG which would be equal to the triangle AEF if, in the descent, the presence of air did not diminish the degree of impetus of the natural gravity. Consequently, the triangle FDG is smaller than the triangle AEF and much smaller than the triangle FAB. Therefore, the distance travelled in the descent and illustrated by the triangle FDG will be less than the distance travelled in the ascent in an equal time and illustrated by the triangle FAB. Moreover, since the ratio of the two triangles FAB and FDG which have an equal height is equal to the ratio of their bases, DG, i.e. the very last impetus acquired at the end of the fall will be smaller than the impetus of projection BA with which the ball began its ascent.

If the obstacle of air was removed completely, the rectangular surface of the impetus ABFE would be equal to the rectangular surface of the impetus EG and the triangle FDG would be equal to the triangle FAB. Therefore, the distances of ascent and of descent in equal times would be equal and the initial impetus of projection AB would be exactly equal to the very last degree of impetus of the descent DG.

The truth of this theory is confirmed by experience. A very heavy ball of lead A is projected upwards with a weak impetus so that the resistance of air is very small. The distances of ascent and of descent in equal times are found to be exactly equal. The impetus at the beginning of the ascent and that at the end of the descent are also equal. Do not say that greater heights are required to make distinct the difference between the ascent and the descent of this projected ball. Indeed, if the upwards distance was twice the downwards distance, this could not be hidden when a small pendulum is used for measuring the equal times.

In a projection inclined on the horizon to half the height which could be attained in an uniform movement, the projectile will rise and travel along a parabolic trajectory.

The same occurs in an upwards movement at an inclination to the horizon (Table 23.​1, Fig. 88). Let a heavy body A be impelled obliquely upwards in an uniform movement with an impetus D so that it can travel in a time T the inclined straight distance GF and is able to rise vertically up to a height GH. I claim that rising ends at the midpoint I of this height and afterwards the body falls along a curved, parabolic route. The straight line FG is divided into equal parts at points B, C, K. From these points perpendiculars are drawn to the subjacent plane FH. They are BL, CM, KN, GI which progressively increase in the same ratio as their squares FB, FC, FK, etc. GI is half the height GH. The body A is impelled in an uniform movement with the impetus D in the inclined direction FG. Thus, in equal times it travels equal distances FB, BC, CK, etc. which, therefore, measure the time of travelling. A, after leaving its position of immobility at F, exerts the force of its gravity which acquires impulses growing equally in equal times. Therefore, instead of travelling through GF, it will descend and pass at perpendicular distances from the horizon which are proportional to the squares of the times of travelling. Thus, if at time FB it is depressed through BL, necessarily at time FC it will descend through CM and at time FK it will fall down to N, and so on. Therefore, the trajectory of the projectile is along the curve FLNI which, as we will show, is a parabola. FR is drawn parallel to GI, KN, etc. From the points L, M, N and I parallels to FG are drawn. They meet FR at points O, P, Q, R. Obviously, the figures BO, CP and GR are parallelograms and, therefore, their opposite sides are equal: OF = BL and OL = AB and so on. Therefore, the ratio of the abscissae OF/FP is equal to the ratio of the squares of their successive counterparts OL and PM. Consequently, the points F, L, M, etc. are on a parabola. Then the straight lines FG and FH are prolonged so that GX = FG and HS = FH. X and S are joined. In the triangle FXS, XS is twice GH. By hypothesis, GH is divided into two equal parts at I. Thus, XS is equal to 4 times GI. Thus, the ratio XS/GI is equal to the ratio of the squares of the distances or times FX and FG. Therefore, point S is on the parabola FI. IH is parallel to FR. FS is divided into two halves at H by GH with which it forms right angles. Thus HI is the axis of the parabola to which FH is regularly attached. The portion GI of the axis above the intersection is equal to IH. Thus FG will be a tangent and I is a point on the axis and the apex of the parabola. Therefore, the end of the ascent of the projectile A travelling over the parabola is not above the vertex I, i.e. not beyond half the total raising GH. After point I the projectile will fall. Q.E.D.

If two heavy bodies are impelled vertically upwards at two velocities, the ratio of the apparent elevations is equal to the ratio of the squares of the impulses or of the times.

A heavy body A impelled vertically upwards along the line CF with an impetus D in a time T arrives at the upper point I of its rising and does not progress further (Table 23.​1, Fig. 89). It appears that the distance CF which would be travelled in the same time in an uniform movement with the impetus D is twice this height CI. The heavy body B also impelled vertically upwards with an impetus E in a time V attains the point L at the maximum of its rising. GL is half the distance GH which it would travel in an uniform movement with an impetus E in a time V. I claim that the ratio of the distances CI/GL is equal to the ratio of the squares of the times T and V, or to the ratio of the squares of the impulses D and E. The ratio of the distances CF and GH travelled in an uniform movement is equal to the product of the ratio of the times T/V and the ratio of the velocities D/E. But CF/GH = FI/HL = CI/GL since FI and HL are the halves of CF and GL. Cl and GL are the apparent elevations of the bodies A and B. Consequently, the ratio of the maximum elevation CI of the body A to the maximum elevation GL of the body B is equal to the product of the ratio of the times T/V and the ratio of the velocities D/E. But the distances travelled in an accelerated movement FI and HL due to the repelling impetus of gravity are subtracted from the distances travelled in an uniform movement. Their ratio FI/HL is equal to the ratio of the squares of the times T and V. Therefore, the ratio of the apparent elevations CI/GL is equal to the ratio of the squares of the times T and V, and thus also to the square of the remaining component of the ratio of the velocities D/E. Q.E.D.

Consequently, if the apparent elevations of two bodies thrown upwards are equal, the impulses on one hand and the times on the other hand are equal since the square roots of the components of the equation of the ratio are also equal. Therefore, whenever the apparent rising CI of a body A is equal to the apparent rising GL of another body B, the impetus D is necessarily equal to the impetus E and the time T is equal to the time V.