## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

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.