ANSWERS. 



/, required by the sound, find the depth s approximately from s = 

 with / = 4 ; with this value of s find / 2 ; hence the time of fall / b with 

 which correct s ; etc. Result : s = 70.4 metres. 



(5) (a) 4 min. ; (b) 1 1/60 ; (c) 30 miles per h. ; (d) after 3 m. 28 s. 



(8) (a) 40,000 ft.; (b) 715.5 ft. per second; (c) i m. 40 s. ; 

 {d) 1600 ft. per second; (e) i m. 12.4 s. and 27.6 s. 



(9) 80 ft. per second. 



(10) (a) t=h/v Q - } (b) h-s = 



Page 60. 



(1) (a) 26,000 ft. per second; (b) 34 m. 48 s. 



(2) It represents a cycloid. 



(4) v^R/(2gR 2 ). If v} 5 2gR, the particle will not fall back. 



(5) Height = >?; time of ascending = ^|f i +-j = time of 



o \ / 



falling back = 34 m. 48 s. ; hence whole time = i h. 9 m. 36 s. 



(6) 7 miles per second. 



Page 63. 



(2) v = 26,000 ft. per second ; /= i h. 25 m. 4.5 s. 



(3) 2S = R(#* + e-v*) t or s = 



Page 65. 



(i) lira #=///, for lim /= oo. 



v* CQS Vg^ ^ ^ sin 



(4) Time of ascent T= I __t3Lir l \ -v ; 



V** X <^ 



height of ascent ^ = ^7 log f i 4- - 2 ) 



PART I 12 



