336 PRACTICAL MATHEMATICS 



M m m M 



and ve m v = ; --- r-e m 



k k 



]cot m / fort \ 



ve m = v ~(e m 1 ) 

 fc \ / 



_ko_t m / _M\ 



and v = v e m f 1 e m J 



as Ka m / K o t \ 



v = -j- = v e m T (l e m } 



dt . K \ J 



Now 



Integrating s = r-2 e m ( t + ^-e m } + Const 



L>fi L* \ l/*rt J 



Kg K \ Kg / 



TtVD TTl^ 



but when 2 = 0, 5=0, 0= r2 + Const 



,2 / Jcat 



_ '' ' II / _ I Iff/ I _ 



and 



l _ e 



r 



At the highest point the velocity is evidently 0, 



Jcgt m , _kgt^ 



and ve m -( 1 e m = 



-M mf - 

 or v e m = ( 1 e m 



K \ 



m 



kv Q + m 

 Jcg m 



m Jcv + m 

 and k0 & e 



and this gives the time taken to reach the highest point. By 

 giving t this value in the expression for s, the vertical distance 

 of the highest point above the point of projection can be de- 

 termined. 



~. m c m~\ f, -M\ m 



Since s = r~ -{ fn H M 1 ~~ * m r ~ ~T~t 



kg I k ) I ) k 



