Motion of Projectiles. 1 95 



4 hy cos a 2 = 4 /i a? sin a cos a a? 2 , 

 gives 



4hy cos a 2 = 8 h a sin a 2 cos a 3 4 ft 2 sin c 2 cos a 2 , 



from which we obtain 



y or D = /i sin a 3 . 



This determines the vertex of the axis, since d y being zero at the 

 point , the tangent at B is parallel to ^C, or perpendicular to 

 BD. 



308. We propose now to determine the direction AZ, to be 

 given to a projectile in order that it may fall upon a known point Fig.160. 

 Jlf, that is, the inclination that a mortar, for instance, must have 

 to throw a shell upon the known point M. 



The perpendicular MP upon the horizontal line passing 

 through the point A, being drawn, the distance #P, and the an- 

 gle MAP, are to be considered as known. AP being designated 

 by c, and the angle MAP by e, we shall have 



c sin e 



cos e : c : : sin e : MP = , 



cos e 



we have, therefore, for the point M, a? = c, and 



csin e 



** " COS 6 



Substituting these values in the equation 



4hy cos a 2 = 4hx sin a cos a a; 2 , 



we obtain, 



4 he sin e cos a 2 



= 4 h c sin a cos a c 2 , 



cos e 



or 



4hsm e cos a 2 = 4 ft sin a cos a cds c c cos e, 

 or 



4 /i os a (sin a cos e sin e cos a) = c cos e. 



Trig. 11, 

 that is, 



4 h cos a sin (a e) = c cos c ; 



