Two points, A and B, are on cliff tops on either side of a deep valley - HSC - SSCE Mathematics Extension 1 - Question 6 - 2009 - Paper 1

We need to show that y₁ = y₂ when x₁ = x₂. Using T from the previous part, we find:

For projectile from A:

$y₁ = U T sin θ - \frac{1}{2}gT²$

For projectile from B:

$y₂ = h - V T sin θ - \frac{1}{2}gT²$

Setting these equal gives:

$U T sin θ - \frac{1}{2}gT² = h - V T sin θ - \frac{1}{2}gT²$

Simplifying, we find:

$U T sin θ + V T sin θ = h$

Factoring out T sin θ:

$(U + V) T sin θ = h$

Since we have expressions for T and equate them, we can conclude the projectiles collide.

Substituting x = λR into the position equations gives:

From A:

$λR = U T cos θ$

From B:

$λR = R - V T cos θ$

Setting these equal:

$UT cos θ = R - V T cos θ$

Rearranging:

$V T cos θ = R - UT cos θ$

Thus,

$V = \frac{R - U T cos θ}{T cos θ}$

We substitute ( T = \frac{R}{(U + V) cos θ} ):

$V = \frac{R - U \frac{R}{(U + V)} }{\frac{R}{(U + V)}}$

This leads to:

$V = \left(1 - \frac{1}{\lambda} \right) U$