512 PHILOSOPHICAL TRANSACTIONS. [aNNO 1721. 



For L is the parameter of the trajectory at a ; hence the solution appears 

 from Prop. 2. 



Scholium. The experiment will be conveniently made, by erecting ag per- 

 pendicular to the horizon, drawing the direction ab bisecting the angle gag, 

 the line ac being parallel to the horizon. For in this case the required altitude 

 is equal to half the distance ac. 



Prop. V. Given the Direction and Velocity of Projection; to Jind the In- 

 tersection of the Trajectory with a Line passing through the Point of Pro- 

 jection. — Let the body be projected from the point a, fig. 4, in the direction 

 AB. Raise ac opposite to the direction of gravity, and equal to the height due 

 to the projectile velocity, and draw ce perp. to ac Make ap = ac, and form- 

 ing the angle baf = the angle bag. Let ak be the line whose meeting with 

 the trajectory is required. Draw fi perp. to ak, and meeting ce in d. Take 

 DE = CD, and draw ek perp. to ce : then will k be the point required. 



For, in fi produced take if = fi, and draw fA, ffi, fk, fk. Because fia is a 

 right angle, and if = fi, hence fA is also := fa. But by construe, af is = ac, 

 and acd is a right angle. Therefore the points c, f, f are in the circle de- 

 scribed with the centre a, and is touched by the line cd. Therefore df, dc, 

 Df are continual proportionals. But de is = cd by constr. therefore df, de, 

 of are continual proportionals; and therefore, because of the common angle 

 at D, the triangles fed, ffio are similar, and the angle fed = the angle FfE. 

 Therefore the three points f, e, f are in the circle which de touches at e. But 

 because fi is = if, and fik is a right angle, the centre of that circle is in the 

 line ik; and because dek is a right angle, that centre is also in ek. There- 

 fore K is the centre of that circle, and conseq. pk is = ek. Now, by Prop. 3, 

 F is the focus of the trajectory, and ga the 4th part of the parameter at the 

 point a. Hence, since ge is perp. to ag and ke, and fk = ek, the point k 

 will be in the trajectory, by conies, a. e. d. 



Prop. VL Having the same Things given ; to find the Meeting of the 

 Trajectory with any Line given in position, — Let the body be projected from 

 the point a, fig. 5, in the direction ab ; and let gh be the line to meet the 

 trajectory. Draw ca in the direction of gravity, and equal to the height due 

 to the given projectile velocity ; also draw af = ag, and making the angle 

 bap = the angle bag; and draw ce perp. to ag. Draw fi meeting gh at 

 right angles in i, and ge in d; and take if = fi. Take de a mean propor- 

 tional between df and Df ; lastly, drawing ek perp. to ge, then k will be the 

 point required, a. e. i. 



Joining Ef, this is demonstrated in the same way as the foregoing. 



