## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

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.