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+//Ball in frictionless tube
+//refer fif. 18.4
+u1=sqrt(2*9.81*2) //m/sec
+u2=0
+//By principle of conservation of momentum
+//v1+2*v2=6.264
+//From defination of coefficient of restitution
+//case(1)-e=1
+//v2-v1=6.264
+//solving
+v2=4.176 //m/sec
+v1=6.264-(2*4.176) //m/sec
+//Let h be the height to which hanging ball will rise
+//Change in K.E=Work Done
+h=(v2^2)/(2*9.81) //m
+//case(2)- e=0.7
+//v2-v1=4.385
+//solving
+bv2=(6.264+4.385)/(3) //m/sec
+//Height to which ball will rise
+h2=(bv2^2)/(2*9.81) //m
+printf("\nCase(1)-\nh=%.4f m\nCase(2)-\nh2=%.4f m ",h,h2)