Question

A lead ball dropped in a lake from a diving board 20 m above the water hit the water with certain velocity and then sinks to the bottom with the same constant velocity. If it reaches the bottom in 6s after it is
dropped, the depth of the lake is (g = 10 m/s2)

Answer 1

Given:

A lead ball dropped in a lake from a diving board 20 m above the water hit the water with certain velocity and then sinks to the bottom with the same constant velocity. It reaches the bottom of the lake in 6 seconds.

To find:

Depth of the lake

Calculation:

Let velocity of the lead ball just before hitting the water of the lake be v :

$\therefore \: {v}^{2} = {u}^{2} + 2gh$

$= > {v}^{2} = {0}^{2} + (2 \times 10 \times 20)$

$= > {v}^{2} = 400$

$= > v = 20 \: m {s}^{ - 1}$

So , this is the terminal velocity with which the ball moves constantly within the water and reaches the bottom of the lake.

Let depth be d :

$\therefore \: d = v \times t$

$= > d = 20 \times 6$

$= > d = 120 \: m$

So final answer is :

Depth of the lake is 120 m.