Two stones are thrown vertically upward from the same point with the same speed of $30$ m/s, the second one $2$ s after the first. At what height above the point of projection do they collide? ($g = 10\ \text{m/s}^2$)
Answer: (B) $40$ m
Let $t$ be the time after the first stone is thrown.
First stone: $h_1 = 30t - 5t^2$. Second stone: $h_2 = 30(t - 2) - 5(t - 2)^2$.
They collide when $h_1 = h_2$:
$$30t - 5t^2 = 30t - 60 - 5t^2 + 20t - 20 \;\Rightarrow\; 20t = 80 \;\Rightarrow\; t = 4\ \text{s}$$
$$h = 30(4) - 5(4)^2 = 120 - 80 = 40\ \text{m}$$
Check: the first stone reached its top at $3$ s and is falling; the second is $2$ s into its rise. By symmetry they meet $1$ s either side of the average time, which is consistent.
Solution by Sreeraj P, M.Sc Physics