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+//soldier fires a bullet
+//refer fig 13.18
+//equation of trajectory of bullet is known thus
+//For the point on ground where bullet strikes
+y=-50  //m
+x=100 //m
+u=31.32  //m/sec
+//alpha=0 or
+alpha=atand(2)  //degree
+//when alpha =0
+//Horizontal component of velocity
+vx=31.32  //m/sec
+//Vertical component of velocity
+vy=sqrt(2*9.81*50)  //m/sec
+//Velocity of strike
+v=sqrt((31.32^2)+(31.32^2))  //m/sec
+theta=atand(1)  //degree
+//when alpha=63.435 degree   vx=14.007 m/sec
+//vy=42.02 m/sec
+bv=sqrt((14.007^2)+(42.02^2))  //m/sec
+btheta= atand(42.02/14.007)  //degree to horizontal
+printf("\nalpha=%.2f degree\nv=%.2f m/sec\ntheta=%.2f degree\nv=%.2f m/sec\ntheta=%.2f degree to horizontal",alpha,v,theta,bv,btheta)
+
+
+
+