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
The motion above the top of the tower is symmetric: the ball takes $1$ s to rise to its highest point and $1$ s to fall back. So its speed at the top of the tower is $v = g \times 1 = 10$ m/s.
From the foot to the top: $v^2 = u^2 - 2gh$:
$$100 = 900 - 20h \;\Rightarrow\; h = 40\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics