Q 11-02-032NEETJEE MainMedium
A ball is thrown vertically upward from the foot of a tower with a speed of $30$ m/s. It passes the top of the tower on its way up and comes back past the top $2$ s later. The height of the tower is ($g = 10\ \text{m/s}^2$)
Answer: (A) $40$ m
Concept. Motion above the tower top is symmetric: $1$ s up and $1$ s down, so the speed at the top is $g \times 1$.
Step 1. Speed at the top. $v = 10$ m/s.
Step 2. Height.
$$v^2 = u^2 - 2gh \;\Rightarrow\; 100 = 900 - 20h \;\Rightarrow\; h = 40\ \text{m}$$
Answer. $40$ m (Option A).
Solution by Sreeraj P, M.Sc Physics