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

To show that the projectiles collide, we need to equate the vertical positions y₁ and y₂ at time T:

From the equations of motion:

• For the projectile from A:

$y₁ = U sin θ - \frac{1}{2} g T²$

• For the projectile from B:

$y₂ = h - V sin θ - \frac{1}{2} g T²$

Setting y₁ = y₂ at time T:

$U sin θ - \frac{1}{2} g T² = h - V sin θ - \frac{1}{2} g T²$

The terms (-\frac{1}{2} g T²) cancel out:

$U sin θ = h - V sin θ$

Rearranging gives:

$h = U sin θ + V sin θ$

This shows that the vertical positions are equal, indicating a collision.

If the projectiles collide on the line x = λR, we can substitute x = λR into the horizontal motion equations:

From the projectile from A:

$x₁ = U cos θ imes T = λR$

For the projectile from B:

$x₂ = R - V cos θ imes T = λR$

Equating both expressions for x:

$U cos θ imes T = λR = R - V cos θ imes T$

Substituting T from part (i):

$U cos θ \left(\frac{R}{(U + V) cos θ}\right) = λR$

Cancelling R and cos θ:

$U = λ(U + V)$

Expanding gives:

$U = λU + λV$

Rearranging to isolate V:

$U - λU = λV$

Factoring out U:

$U(1 - λ) = λV$

Hence, solving for V gives:

$V = \left(\frac{1}{λ - 1}\right)U.$