#### Answer

The lake is 19.7 meters deep.

#### Work Step by Step

We can find the speed when the ball hits the water.
$v^2 = v_0^2+2ay = 0+2ay$
$v = \sqrt{2ay}$
$v = \sqrt{(2)(9.80~m/s^2)(5.0~m)}$
$v = 9.9~m/s$
We can find the time it takes the ball to drop 5.0 meters through the air.
$y = \frac{1}{2}gt^2$
$t = \sqrt{\frac{2y}{g}}$
$t = \sqrt{\frac{(2)(5.0~m)}{9.80~m/s^2}}$
$t = 1.01~s$
Since the ball reaches the bottom 3.0 seconds after it is released, the ball falls through the water for 1.99 seconds. We can find the depth $d$ of the lake.
$d = v~t$
$d = (9.9~m/s)(1.99~s)$
$d = 19.7~m$
The lake is 19.7 meters deep.