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20 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—

$$Energy = \frac{5(20)^2}{19.6} = \frac{2000}{19.6} = 102.04$$ nearly.

Again, let it be required to find the height to which the mass of the last question will ascend before it stops. We know that its energy is 102.04, and that its mass is 5. Dividing 102.04 by 5, we obtain 20.408 as the height to which this mass of five kilogrammes must ascend in order to do work equal to 102.04 kilogrammetres.

30. In what we have said we have taken no account either of the resistance or of the buoyancy of the atmosphere; in fact, we have supposed the experiments to be made in vacuo, or, if not in vacuo, made by means of a heavy mass, like lead, which will be very little influenced either by the resistance or buoyancy of the air.

We must not, however, forget that if a sheet of paper, or a feather, be shot upwards with the velocities mentioned in our text, they will certainly not rise in the air to nearly the height recorded, but will be much sooner brought to a stop by the very great resistance which they encounter from the air, on account of their great surface, combined with their small mass.

On the other hand, if the substance we make use of be a large light bag filled with hydrogen, it will find its way