44-46] PARABOLIC MOTION. 45 



Thus, after an interval measured by F sin a/g, y vanishes, and 

 the particle has no velocity parallel to the axis y, it is therefore 

 moving parallel to the axis x. Previously to this it had a velocity 

 in the positive direction of the axis y, and after this it has a 

 velocity in the negative direction of the same axis. Its path 

 therefore has a vertex, which is reached after an interval 



Vsma/g, = say, 

 from the beginning of the motion. 



If we refer the motion to parallel axes of # , y (y f being 

 positive in the opposite sense to y} through the vertex A, and 

 take t to measure the time of moving from the vertex A to any 

 point P we shall have 



_ = 0. with -7-7 = V cos a, and x 0, at time tf = 0, 

 dt z at 



and = g&amp;gt; wi th = 0, and y = 0, at time t = 0. 



Cut Cut 



Hence x = V cos at , y 1 = \gP. Eliminating t , we have 

 , a 2 F 2 cos 2 a , 



* 2 =- - -- -y, 



so that the path of the particle is a parabola with vertex at A. 



We might have deduced this result analytically from the equations #=0, 

 y= g. Integrating and determining the constants so that when =0, x=x^ 

 x Fcosa, and y=y, y Fsina, we find 



X=X Q + Fcosa. t, 



Eliminating t we have 



F 2 sin 2 a 



the equation of a parabola whose axis is parallel to the axis y, and whose 

 vertex is at the point 



F 2 sin a cos a F 2 sin 2 a 



The theorem of this Article was discovered by Galilei. 



46. Examples. [In these examples the axis y is supposed to be the 

 vertical at a place.] 



1. Write down the length of the latus rectum of the above parabola. 



2. Show that the height of the directrix above the starting point is F 2 /2#. 



3. If v is the velocity at any point of the path show that the point is at a 

 distance iP/Zg below the directrix. 



