Question

A lead ball is dropped into a lake from a diving board 5.0 m

above the water. After entering the water, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the water. The ball reaches the bottom 3.0 s after it is released. How deep is the lake?

Short answer

The lake is$19.7m$deep.

Step-by-step solution

Given data

Height of the diving board, $h=5m$

The ball reaches the bottom in time role="math" localid="1648239943091" $T=3s.$

We have to find the depth $d$of the lake.

Calculation of time taken by the ball to drown.

If we consider the free fall of the lead ball then time needed by the ball to reach the surface of water can be obtained by,

$h=\frac{1}{2}g{t}^2$

Therefore, the time needed to reach the surface of the lake,

role="math" localid="1648239398400" $t=\sqrt{\frac{2h}{g}}$

Now, the speed of the ball while reaching the water surface is, $v=\sqrt{2gh}\left[as{v}^2=0+2gh\right]$

Time needed by the ball to drown into the lake from the surface,

$$\begin{gathered} t\text{'}=\frac{d}{v} \\ t\text{'}=\frac{d}{\sqrt{2gh}} \end{gathered}$$

Calculation of depth

Total time needed to drown the ball into the lake is, $T=t+t\text{'}$

Therefore,

role="math" localid="1648240364972" $$\begin{gathered} T=\sqrt{\frac{2h}{g}}+\frac{d}{\sqrt{2gh}} \\ \frac{d}{\sqrt{2gh}}=T-\sqrt{\frac{2h}{g}} \\ d=T\sqrt{2gh}-2h \\ d=3\sqrt{2\times 9.8\times 5}-2\times 5\left[Takeg=9.8m/s^2\right] \\ d=29.7-10 \\ d=19.7m \end{gathered}$$