346 PRACTICAL MATHEMATICS 



dy_ 

 dx 



(14) Vl x 2 -j^ + y = 1, given y = 2, when a; = 



(15) -T-+ y tan # = sec x, given ?/ = 0, when x = 

 dx 



(16) -T-+ y tan # = sin 2 2#, given / = ; when x = 



UM? 



dit ^ 



(17) -g + ?/ = # 2 , given */ = 0, when a? = 



(18) Show that the differential equation -^ + yx m = y n x v 



dz 



reduces to the form -r- + (1 - n)z x m = (1 - n)ce p if z = y l ~ n 

 dx 



(19) Apply the result of Question 18 to solve the differential 



equation a? -^- + x z y - y 3 subject to the condition that y = 1 

 dx 



when x = 1. 



(20) Plot the values of s and t given in paragraph 167 on 

 squared paper between t = 20 and t = 28. Use the graph to find 

 the time of flight necessary for a horizontal range of 40,000 ft. 

 What is the angular elevation at which the projectile must be 

 fired to give this range ? 



(21) The projectile in paragraph 167 is given an elevation of 



a 



tan -1 - ; that is, for the horizontal muzzle velocity of 2700 ft. 

 y 



per sec., the vertical velocity must be 600 ft. per sec. Find the 

 horizontal and vertical components of its velocity, 36 seconds 

 after projection. What is the magnitude of the velocity at this 

 instant and in what direction is it travelling ? 



(22) A body of mass 5 Ib. is projected upwards in a resisting 

 medium with an initial velocity of 200 ft. per sec. ; the resistance 

 of the medium being kv Ib., where v is the velocity of the body 

 and A; is a constant. If the body takes 3-5 seconds to reach its 

 highest point find the value of k (k lies between 0-04 and 0-05). 

 What is the greatest height to which the body will rise and what 

 will be the velocity of the body 2 seconds after projection ? Take 

 g = 32-2 ft. per sec. 2 . 



(23) A body of mass 5 Ib. is projected upwards in a resisting 

 medium with an initial velocity of 200 ft. per sec. ; the resistance 

 of the medium being kv 2 Ib., where v is the velocity of the body 

 and A; is a constant. If the body takes 3-5 seconds to reach its 

 highest point, find the value of k (k lies between 0-0004 and 

 0-0005). What is the greatest height to which the body will 

 rise; and what will be the velocity of the body 2 seconds after 

 projection ? Take g = 32-2 ft. per sec. 2 . 



