Two particles: different start times and heights

5 A particle is projected vertically upwards with speed \(40 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) alongside a building of height \(h \mathrm {~m}\).

1. Given that the particle is above the level of the top of the building for 4 s , find \(h\).
2. One second after the first particle is projected, a second particle is projected vertically upwards from the top of the building with speed \(20 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Denoting the time after projection of the first particle by \(t \mathrm {~s}\), find the value of \(t\) for which the two particles are at the same height above the ground.

Two particles \(A\) and \(B\) are released from rest from different starting points above a horizontal surface. \(A\) is released from a height of \(h\) metres. \(B\) is released at a time \(t\) seconds after \(A\) from a height of \(kh\) metres, where \(0 < k < 1\) Both particles land on the surface \(5\) seconds after \(A\) was released. Assuming any resistance forces may be ignored, prove that $$t = 5(1 - \sqrt{k})$$ Fully justify your answer. [5 marks]