280
THEORY" AND PRACTICE
TRACT 37,
a rampart, a block of wood, &c, the ball will lodge in itwithout any sensible recoil; so that, in the present considera-tion, if the body be elastic, its elasticity may be neglected.
136. When a ball is discharged against such an obstacle,it will not only make an impression, but will penetrate to acertain depth ; and as this cannot happen without tile ballsuffering a great resistance from the obstacle, it will have itsmotion gradually diminished, and finally quite destroyed.To find the depth to which the.ball can penetrate, we mustdetermine the resistance it suffers while it penetrates into theobstacle; for, whatever matter the obstacle consists of, whe-ther wood, or earth, See, a greater cavity will always requirea greater force; and since the elasticity is not considerable,the ball will meet with the same resistance nearly, after en-tering a certain depth, as at the beginning of its penetra-tion; excepting that, in such cases, the farther the penetra-tion is continued, the resistance will be rather increased bythe greater accumulation and condensation of the parts ofthe obstacle in the front of the striking body. The resist-ance is therefore nearly a constant and uniform force, noway depending on the velocity; and so will be nearly simi-lar to the action of gravity, bv which it happens that a body,projected directly upwards, loses equal quantities of its mo-tion in equal times, whether the velocity be quick or slow.The quantity of this force depends on the strength of theobstacle, and the w idth of the cavity made by the ball, whichis proportional to the square of the ball’s diameter. Hencethe calculations for the circumstances of this motion and re-sistance, considering this last as a constant quantity, will besimilar to the resisting force of gravity on a body projectedupwards, or to any motion whatever resisted by a force thatis constant.
137. Therefore, lety denote the constant resisting force,or strength of the wood, or other matter; 5 the space ordepth penetrated ; v the first velocity, and t the time of pe-netration ; also#" = 16 t ’ t feet, or 16 feet only, the space a