Q 11-02-029NEETJEE MainHard
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
Concept. Write both heights as functions of the time since the first throw and set them equal.
Step 1. Heights. $h_1 = 30t - 5t^2$, $h_2 = 30(t - 2) - 5(t - 2)^2$.
Step 2. Solve.
$$30t - 5t^2 = 30t - 60 - 5t^2 + 20t - 20 \;\Rightarrow\; 20t = 80 \;\Rightarrow\; t = 4\ \text{s}$$
Step 3. Height. $30(4) - 5(16) = 40$ m.
Answer. $40$ m (Option B).
Solution by Sreeraj P, M.Sc Physics