46 MOTION IN TERMS OF ACCELERATION. [CHAP. IV. 



4. Prove that the time until the particle is again in the horizontal plane 

 through the point of projection is 2 Fsin a/g. [This is called the time of flight 

 on the horizontal plane through the point of projection.] 



5. Prove that the distance from the starting point of the point where the 

 particle strikes the horizontal through the starting point is F 2 sin 2a/g. [This 

 is called the range on the horizontal plane through the point of projection.] 



6. To find the range and time of flight on an inclined plane through the 

 point of projection. Let 6 be the inclination of the plane to the horizon. 



Kesolve up the plane, and at right angles to it. The resolved accelerations 

 are 



Fig. 31. 



the resolved initial velocities are 



Fcos(a-0), Fsin (a- 6); 



the resolved velocities at time t are 



V cos (a - 6} - gt sin 6, Fsin (a - 6} - gt cos 6 ; 



the distances described in time t parallel and perpendicular to the inclined 



plane are 



Vt cos (a - 0) - \g$ sin &amp;lt;9, Vt sin (a - 6} - \g$ cos 6. 



The time of flight is obtained by making the second of these equal to zero, it is 



2 Fsin (a- &amp;lt;9) 

 gcosd 



The range is found by substituting this value for t in Vt cos (a - &) - \gfi sin 6. 

 Prove that the range in question is 



and that this is the same as 

 72 



[sin(2a-0)-sin0]. 



g cos 2 6 



7. Prove that, when the velocity of projection is given, the range on an 

 inclined plane is greatest when the direction of projection bisects the angle 

 between the plane and the vertical. 



8. Show that if a parabola is constructed having its focus at the point of 

 projection &amp;gt;S , its axis vertical, and its vertex at a height V 2 /2g above the point 





