When a ball is dropped into a lake from a height 4.9 m above the water level, it hits the water with a velocity v and then sinks to the bottom with the constant velocity v. If it reaches the bottom of the lake 4.0 s after it is dropped, the approximate depth of the lake is

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(2) 29.4 m. Fall in air: 4.9 = 1/2×9.8× t² ⇒ t = 1 s, and v = 9.8×1 = 9.8 m/s. Remaining time in water: 4.0 - 1 = 3 s at a constant 9.8 m/s. Depth = 9.8×3 = 29.4 m.

When a ball is dropped into a lake from a height 4.9 m above the water level, it hits the water with a velocity v and then sinks to the bottom with the constant velocity v. If it reaches the bottom of the lake 4.0 s after it is dropped, the approximate depth of the lake is

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