A projectile problem

A boy stands on the edge of a 100m high cliff and shoots a projectile (A) vertically upwards at 30 ms^-1. One second later his friend at the base of the cliff shoots a second projectile (B) vertically upwards at 50 ms^{- 1}.

Determine:
(a) the height above the ground at the base of the cliff where the two projectiles pass each other
(b) the time after the projectile A was launched that this happens
(c) the velocities of projectile A and B when they pass each other

Let the time after projectile A is launched until they pass each other be t and let the height above the ground when this happens be h.

Take g = 9.8 ms^{- 2}.

For projectile A: h = 100 + 30t + ½ gt² equation 1
For projectile B: h = 50(t-1) + ½ g(t-1)² equation 2
Therefore: 100 + 30t + ½ gt² = 50(t-1) + ½ g(t-1)² = 50t – 50 + ½ x 9.8 x (t² – 2t +1)
So remembering that g will be negative if the velocities upwards are positive:
100 + 30t + ½ gt² = 59.8t – 54.9 + ½ gt²
154.9 = 29.8t and so t = 5.2 s

Substituting for t in equation 1 gives: h = 100 + 30x5.2 – ½ x9.8x5.2²
h = 255.91 - 132.39 = 123.5 m

The projectiles pass each other when they are 123.5 m above the ground

Checking this using equation 2 also gives h = 123.5 m

Velocity of A at h = 123.5 m

v = u –gt = 30 – 9.8x5.2 = -20.1 ms^-1 A is travelling downwards

Velocity of B at h = 123.5 m

v = u –gt = 50 – 9.8x4.2 = +8.84 ms^-1 B is travelling upwards