When a bullet is fired from a gun with its barrel directed towards the target, it starts falling downwards on account of acceleration due to gravity.
Due to which the bullet hits below the target. Just to avoid it, the barrel of the gun is lined up little above the target, so that the bullet after travelling in parabolic path hits the distant target.

Yes, the flight of a bird is an example of composition of vectors. As the bird flies, it strikes the air with its wings W, W along WO. According to Newton's third law of motion, air strikes the wings in opposite directions with the same force in reaction. The reactions are OA and OB. From law of parallelogram vectors, OC is the resultant of OA and OB. This resultant upwards force OC is responsible for the flight of the bird.

Hatrick knew that the bullet falls under gravity, when fired. It follows a parabolic path. Peterson was firing the bullet by keeping the gun in the line of the sight and hence missing the target. Thus, the bullet has to be fired by keeping the gun tilted above the line of sight to hit the target.

The horizontal range of the projectile is given by 
R = \(\frac { { u }^{ 2 }sin2\theta }{ g } \)
If angle of projection 
\(\theta ={ 45 }^{ o }-\alpha \quad R={ R }_{ 2 }\)
\({ R }_{ 2 }=\frac { { u }^{ 2 }sin2({ 45 }^{ o }-\alpha ) }{ g } =\frac { { u }^{ 2 } }{ g } cos2\alpha \) 

(i) The horizontal range of the projectile is given by 
R = \(\frac { { u }^{ 2 }sin2\theta }{ g } \)
If angle of projection \(\theta ={ 45 }^{ o }+\alpha \quad \)and \(R={ R }_{ 1 }\)
Then, \({ R }_{ 1 }=\frac { { u }^{ 2 }sin2({ 45 }^{ o }+\alpha ) }{ g } =\frac { { u }^{ 2 } }{ g } cos2\alpha \) 
(ii) R = \(\frac { { u }^{ 2 }sin2\theta }{ g } \)
If angle of projection 
\(\theta ={ 45 }^{ o }-\alpha \quad R={ R }_{ 2 }\)
\({ R }_{ 2 }=\frac { { u }^{ 2 }sin2({ 45 }^{ o }-\alpha ) }{ g } =\frac { { u }^{ 2 } }{ g } cos2\alpha \)