ON LAGRANGE’S BALLISTIC PROBLEM. 
211 
Let s, denote the value of s at A. Then at A we have 
s = Si> v = (f±ff, 
t' — C A + cA 2 + C 3^A 3 + • • • ) 
where S A stands for (s A — S 2 )/S 2 . 
Along AC we have the formulae already written for t' and dt'/ds, and we have further 
v = C-Aj ( 1 - 20 £+ 90 ^- 140 f 3 + 7 °O> i = 
|^-5 - = (: s '~' s 2/ 7 ’ + ,9 ^ (20-l 8 0^+420^-280r)> 
Or (t 1 - ' b 
(r' + s? 
and we have to put 
ds = t a dS, dr = S 2 (B / 1 + 2B / A+3B / 3 (f + ...) dS. 
The limits of integration are <i A and S', which is (s'—S 2 )/S 3 . 
The value of S A is to be found by reversing the series 
r-S 2 = (s—S 3 ) {B', + (BVS a ) (,-S.) + (B' s /S/) (s-S.,) 2 + ...}, 
and putting r' for r and s A for 5. If we write e' for (/ —R,)/S 2 the result is 
? __ _£_ [ ’ 2 /2 , 
A Bh B? e 
f2B? 
B'a 1 
/5B? 
IB? 
B?J 
VBV 
+ 
BT B? 
/4 
, BV B' 2 2 B' B'» B' 2 B', Ru 
+ (14 B7 “ 1 TW +3 bv + 6 T?T _ bv ; / 
Thus S x is known in terms of r'. 
37. Formula for the Time .—We work out the formula for t' or t — T 2 in terms of S, 
or (s — S 2 )/S 2 , and e, or (r— R 2 )/2h, at any point answering to simultaneous values of 
r and s which can occur in the second reflected wave from the left. For this we first 
perform the integrations with respect to r and s and then suppress the accents on S' 
and e. We record the results as far as terms of the fourth order. 
The terms of the first order present themselves in the form 
/ 1 + e + b v \ 5 
\ 1 + e + cf / 
A a + 
(l + e + d) 
; 5 \c\ (S—S A ), 
and it is simpler to leave the factors 
/ f + e + (1 a V' 1 
\1 +e+S/’ (l+e + S) 5 
