Question: 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 = a (b - \sqrt{k}) where $a$ and $b$ are integers to be found.

1. **State the problem:** Two particles A and B are dropped from rest from heights $h$ and $kh$ respectively, with B released $t$ seconds after A. Both hit the ground 5 seconds after A's release. We need to prove that $$t = a(b - \sqrt{k})$$ where $a$ and $b$ are integers. 2. **Recall the formula for free fall:** The distance fallen under gravity $g$ in time $T$ is given by: $$s = \frac{1}{2} g T^2$$ 3. **Apply to particle A:** Particle A falls from height $h$ and hits the ground after 5 seconds, so: $$h = \frac{1}{2} g (5)^2 = \frac{1}{2} g \times 25 = \frac{25}{2} g$$ 4. **Apply to particle B:** Particle B is released $t$ seconds after A, so it falls for $(5 - t)$ seconds before hitting the ground. Its height is $kh$, so: $$kh = \frac{1}{2} g (5 - t)^2$$ 5. **Substitute $h$ from step 3 into step 4:** $$k \times \frac{25}{2} g = \frac{1}{2} g (5 - t)^2$$ 6. **Simplify by canceling $\frac{1}{2} g$ on both sides:** $$k \times 25 = (5 - t)^2$$ 7. **Take the square root of both sides:** $$\sqrt{k \times 25} = 5 - t$$ $$5 \sqrt{k} = 5 - t$$ 8. **Solve for $t$:** $$t = 5 - 5 \sqrt{k}$$ 9. **Rewrite $t$ in the form $a(b - \sqrt{k})$:** $$t = 5(1 - \sqrt{k})$$ So, $a = 5$ and $b = 1$. **Final answer:** $$\boxed{t = 5(1 - \sqrt{k})}$$