178 
MINUTES OF PROCEEDINGS OF 
“ equations of motion” correctly, as in the former case of a projectile 
in vacuo. 
They are— 
horizontal acceleration —f cos </> =f ( j / ^ 
CIS 
vertical 
dP 
dht 
W 
dt ds * 
„ . , „dy „dy dt 
m 
.© 
A simpler form can be obtained before substituting the value of f. 
Multiply (k) by f ~ and ( l) by and subtract. 
dh/ dx d 2 x dy 
W Jt 
dx 
dt 2 dt ^ dt* 
Eq» (j). 
Eq« (i). 
dp dx _ 
dt dt 
.( 1 ) 
d 2 x 
dsr dx 
ds\ dx 
(*) beoomes 3?.:— 25 iv i = - 22 it • i 
dx 3 
In order to integrate this, bring ^ over to the left and multiply by 
Eqn (1). —. = 
L _ _ 1 dp 
dx q dt 
dt 
d 2 x 
, ~d& 2b a _ 
+p) Tt’ 
dt 
,1 , 1 , U / , f\ 
~ 3 -Sp = - 3 -v + 7V + s)' 
dt\ 
by integrating, the constant being determined by putting 0 = 0 (at the 
dx 
vertex), so that p = 0; = %. 
1 1 25 o\ 
^ n +/)• 
Hence , putting for - its value from (1), viz.-- 
q (It 
.( 2 ) 
2to 0 8 
(3jP +i> 8 ) | 3 ' 
§ 2 - 
= [ 
«& »0 l ^ 
, 25 w n 3 Mfr. , C/ n = retardation at vertex/ 
But -/ = Mg' Where | ir= mass of projectile, 
= y/ suppose. (m) 
rhn 
.(3) 
.*. (integrating) t 
Mg 
resistance at vertex 
weight of projectile 
_ % * k 
{i — r (3? +i) 3 )} 4 ’ 
* Compare definition of no, above. 
