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 \(v\). It reaches the bottom of the lake \(4.0,s\) after it is dropped. The approximate depth of the lake is ............ \(m\)

Pick an option to check your answer.

Verified Solution

Answer & Solution

Correct Answer

(B) \(29.4\)

Detailed explanation

\(v = \sqrt{2gh}\)

\(v = \sqrt{2 \times 9.8 \, m/s^2 \times 4.9 \, m}\)

\(v = 9.8 \, m/s\)

\(t_1 = \frac{v}{g}\)

\(t_1 = \frac{9.8 \, m/s}{9.8 \, m/s^2}\)

\(t_1 = 1.0 \, s\)

\(t_2 = T_{total} - t_1\)

\(t_2 = 4.0 \, s - 1.0 \, s\)

\(t_2 = 3.0 \, s\)

\(h_{lake} = v \times t_2\)

\(h_{lake} = 9.8 \, m/s \times 3.0 \, s\)

\(h_{lake} = 29.4 \, m\)

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