i68.] 



PLANE MOTION. 



uniformly accelerated. This is otherwise directly evident from 

 the nature of the problem. 



Eliminating t between the expressions for x and y, we find 

 the equation of the path 



cose 



(7) 



which represents a parabola passing through the origin. To 

 find its vertex and latus rectum, divide by the coefficient of x % 

 and rearrange : 



2V f 



y\ 



completing the square in x, the equation can be written in the 

 form 



.2 



-^ sm 2 e 



*g J g 



Y= -^cos 2 * (y - ^sin?e\ (;' 



J pr V 2 J 



V* 



ie co-ordinates of the vertex are therefore <*=-^-sin2e, 



V 2 2 V 2 & 



? = sin 2 e; the latus rectum 4 a = - cos 2 e ; the axis is 



O <3 



:ical, and the directrix is a horizontal line at the distance 



v 2 



cos 2 e above the vertex. 



168. Exercises. 



(1) Show that the velocity at any time is v = V?'o 2 2gy. 



(2) Prove that the velocity of the projectile is equal in magnitude 

 to the velocity that it would acquire by falling from the directrix : (a) at 

 the starting point, (b) at any point of the path (see Art. 113). 



(3) Show that a body projected vertically upwards with the initial 

 velocity V Q would just reach the common directrix of all the parabolas 

 described by bodies projected at different elevations with the same 

 initial velocity V Q . 



