Question 166 of 308

When a ball is dropped into a lake from a height $4.9m$ 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 $4.0s$ after it is dropped. The approximate depth of the lake is

A$39.2m$B$19.6m$C$73.5m$D$29.4m$

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Answer: (D) $29.4m$

Using $s=ut+\dfrac{1}{2}a{t}^{2}$
Initially, $u=0$, so time taken to reach from $A$ to $B$ is $t=\sqrt{\dfrac{2s}{g}}=\sqrt{\dfrac{2\times 4.9}{9.8}}=1s$
Now, using ${v}^{2}={u}^{2}+2as$, we have
$v=\sqrt{2gh}=\sqrt{2\times 9.8\times 4.9}=9.8m{s}^{-1}$
$∵{t}_{2}=\left(4-1\right)s=3s$
$∴d={t}_{2}\times v=3\times 9.8=29.4m$