(6)
This is a parabola with vertical axis, of latus-rectum 2u02/g. The range on a horizontal plane through O is got by putting y = 0, viz. it is 2u0v0/g. we denote the resultant velocity at any instant by ṡ we have
ṡ2 = ẋ2 + ẏ2 = ṡ02 − 2gy.
(7)
Another important example is that of a particle subject to an acceleration which is directed always towards a fixed point O and is proportional to the distance from O. The motion will evidently be in one plane, which we take as the plane z = 0. If μ be the acceleration at unit distance, the component accelerations parallel to axes of x and y through O as origin will be −μx, −μy, whence
| d2x | = −μx, | d2y | = − μy. |
| dt2 | dt2 |
(8)
The solution is
x = A cos nt + B sin nt, y = C cos nt + D sin nt,
(9)