Expand upon conceptual example 5. (water balloon)

a boy is hanging in a tree and another boy is on a hill that is lower than the tree. the boy on the hill aims a water balloon gun upward at the boy hanging in the tree. to dodge the water balloon, the boy in the tree lets go of the limb and falls to the ground. if the boy on the hill aims exactly at the boy in the tree and fires right when the boy lets go, prove that the water balloon still hits the boy.

Suppose that the hill is a meters high and the tree is b meters high (a < b), the hill is located c units away from the tree, the balloon is launched at a speed of v, and that the water balloon is launched at an angle of θ with respect to the horizontal.

The horizontal and vertical components of the water balloon are:
(i) x(balloon)(t) = vt*cosθ
(ii) y(balloon)(t) = (1/2)gt^2 + vt*sinθ + a,

while the horizontal and vertical components of the boy are:
(i) x(boy)(t) = c
(ii) y(boy)(t) = (1/2)gt^2 + b.

Now, the water balloon, if it does hit the boy, will hit the boy when the balloon travels c meters to the tree. So:
vt*cosθ = c ==> t = c/(v*cosθ).

At this point, we need to show that the values of y(t) are the same at t = c/(v*cosθ).

With t = c/(v*cosθ):
(i) y(balloon) = (1/2)g[c/(v*cosθ)]^2 + v[c/(v*cosθ)]sinθ + a
= (1/2)gc^2/(v^2*cos^2θ) + vc*sinθ/(v*cosθ) + a
= (1/2)gc^2/(v^2*cos^2θ) + a + c*tanθ.
(ii) y(boy) = (1/2)g[c/(v*cosθ)]^2 + b
= (1/2)gc^2/(v^2*cos^2θ) + b.

The only difference between these two expressions is that we have a + c*tanθ in y(balloon) and b is y(boy).

If we draw out a diagram and use SOHCAHTOA:
tanθ = (b - a)/c ==> b = a + c*tanθ.

This shows that:
y(balloon) = (1/2)gc^2/(v^2*cos^2θ) + a + c*tanθ
= (1/2)gc^2/(v^2*cos^2θ) + b
= y(boy).

Therefore, the balloon hits the boy falling from the tree as required.