A hunter aims his gun and fires a bullet directly at a monkey in a tree. At the instant the bullet leaves the barrel of the gun, the monkey drops. Will the bullet hit the monkey. Substantiate your answer with proper reasoning. 

Given: Let the monkey stationed at A be fired at by a gun from O with velocity u at an angle θ with the horizontal direction OX. Draw AC perpendicular to OX. Let the bullet cross the vertical line AC at B after time t, and let the coordinates of B be (x, y) with respect to origin O.

∴ t = OC / (u cos θ) = x / (u cos θ)     ...(i)

In ΔOAC, AC = OC tan θ = x tan θ     ...(ii)

Clearly, CB = y is the vertical distance travelled by the bullet in time t.

Taking motion of the bullet from O to B along the Y-axis:

y₀ = 0, y = y, uᵧ = u sin θ, aᵧ = −g, t = t

Using the equation of motion:

y = y₀ + uᵧt + (1/2)aᵧt²

⇒ y = u sin θ t − (1/2)g t²

Now, AB = AC − BC = x tan θ − y

⇒ AB = x tan θ − (u sin θ t − (1/2)g t²)

⇒ AB = x tan θ − (u sin θ (x / (u cos θ)) − (1/2)g t²)     [from (i)]

⇒ AB = x tan θ − x tan θ + (1/2)g t² = (1/2)g t²

It means the bullet will pass through point B on the vertical line AC at a vertical distance (1/2)g t² below point A.

The distance through which the monkey falls vertically in time t is also (1/2)g t² = AB.

Conclusion: The bullet and the monkey will pass through point B simultaneously. Therefore, the bullet will hit the monkey.