# Chapter 2 - Kinematics in One Dimension - Exercises and Problems - Page 66: 51

The lake is 29.7 meters deep.

#### Work Step by Step

We can find the speed when the ball hits the water; $v^2 = v_0^2+2ay = 0+2ay$ $v = \sqrt{2ay} = \sqrt{(2)(9.80~m/s^2)(5.0~m)}$ $v = 9.9~m/s$ We use this speed to find the depth $d$ of the lake: $d = v~t = (9.9~m)(3.0~s)$ $d = 29.7~m$ The lake is 29.7 meters deep.

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