Find the trajectory of a cannon ball fired by a cannoneer at 
above his eye level with an initial speed 
.
Ans.
Imagine you are in his place, like the picture below:
Then, suppose you are opening fire at such positive 
-axis direction as that follows along the meridian. In addition, assume that the earth is to be lying flat beneath the cannon ball during its flight, that every pole (
-axis) upheld would be so much right-angled (/normal /perpendicular) to the ground as parallel (-transported) anywhere.
We neglect air friction.
1. Along the 
-direction, the speed is kept 
.
2. Along the -direction, the speed is initially 
. By the laws of gravity, the cannon ball will experience a net force 
due to gravitational pull by the Earth. If upward direction is taken the positive sign, an equation of motion due to the Galilean transformation (i.e., 
) will depict that 
.
3. One another equation 
depicts how distance 
(i.e., the magnitude of displacement 
) varies with time 
during linear motion in constant acceleration 
.
One might have already noticed I am setting the original question aside, as it were. Let’s pinpoint the answer now.
The trajectory should be obtained in the form:
.
where
and
