"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 :"

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"

${t_1} = \sqrt {{{2h} \over g}}$

$= \sqrt {{{2 \times 4.9} \over {9.8}}} = 1\,s$

$\Delta t = 4 - 1 = 3\,s$,

$v = \sqrt {2gh} = \sqrt {2 \times 9.8 \times 4.9} = 9.8$ m/s

$\therefore$ depth $= 9.8 \times 3 = 29.4$ m

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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 :

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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 :

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