Now, it is well known to those who have studied the laws of motion, that a body, shot upwards with the velocity of 9·8 metres in one second, will be brought to rest when it has risen 4·9 metres in height. If, therefore, it be a kilogramme, its upward velocity will have enabled it to raise itself 4·9 metres in height against the force of gravity, or, in other words, it will have done 4·9 units of work; and we may imagine it, when at the top of its ascent, and just about to turn, caught in the hand and lodged on the top of a house, instead of being allowed to fall again to the ground. We are, therefore, entitled to say that a kilogramme, shot upwards with the velocity of 9·8 metres per second, has energy equal to 4·9, inasmuch as it can raise itself 4·9 metres in height. 27. Let us next suppose that the velocity with which the kilogramme is shot upwards is that of 19·6 metres per second. It is known to all who have studied dynamics that the kilogramme will now mount not only twice, but four times as high as it did in the last instance--in other words, it will now mount 19·6 metres in height. Evidently, then, in accordance with our principles of measurement, the kilogramme has now four times as much energy as it had in the last instance, because it can raise itself four times as high, and therefore do four times as much work, and thus we see that the energy is increased four times by doubling the velocity. Had the initial velocity been three times that of the first instance, or 29·4 metres per second, it might in like manner be shown that the height attained would have been 44·1 metres, so that by tripling the velocity the energy is increased nine times. 28. We thus see that whether we measure the energy of a moving body by the thickness of the planks through which it can pierce its way, or by the height to which it can raise itself against gravity, the result arrived at is the same. _We find the energy to be proportional to the square of the velocity_, and we may formularize our conclusion as follows:-- Let _v_ = the initial velocity expressed in metres per second, then the energy in kilogrammetres = _v_²/19·6. Of course, if the body shot upwards weighs two kilogrammes, then everything is doubled, if three kilogrammes, tripled, and so on; so that finally, if we denote by _m_ the mass of the body in kilogrammes, we shall have the energy in kilogrammetres = _mv_²/19·6. To test the truth of this formula, we have only to apply it to the cases described in Arts. 26 and 27. 29. We may further illustrate it by one or two examples. For instance, let it be required to find the energy contained in a mass of five kilogrammes, shot upwards with the velocity of 20 metres per second. Here we have _m_ = 5 and _v_ = 20, hence-- Energy = 5(20)²/(19·6) = 2000/(19·6) = 102·04 nearly.