Q 11-02-098JEE MainJEE Main 2022 (27 Jun, Shift 2)Easy
When a ball is dropped into a lake from a height $4.9$ m above the water level, it hits the water with a velocity $v$ and then sinks to the bottom with the constant velocity $v$. It reaches the bottom of the lake $4.0$ s after it is dropped. The approximate depth of the lake is
Answer: (D) $29.4$ m
Free fall through $4.9$ m:
$$t_1 = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2\times4.9}{9.8}} = 1\ \text{s}, \qquad v = gt_1 = 9.8\ \text{m s}^{-1}$$
In the water it moves at constant $v$ for the remaining $3$ s:
$$\text{depth} = 9.8\times3 = 29.4\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics